MATH 121 - COLLEGE ALGEBRA -
Equations and properties of conic
sections
Question Bank - Set 3
Liberty University
Question 1
Question
Solve the following equation for yin terms of x:
(x−3)2+(y−1)2
4= 1.
Solution
Step 1: Rewrite the equation in standard form:
(x−3)2+(y−1)2
4= 1
(x−3)2+(y−1)2
22= 1.
Step 2: To isolate the term with y, subtract (x−3)2from both sides:
(y−1)2
4= 1 −(x−3)2.
Step 3: Multiply both sides by 4 to eliminate the fraction:
(y−1)2= 4(1 −(x−3)2).
Step 4: Take the square root of both sides to solve for y:
y−1 = ±√4(1 −(x−3)2).
Step 5: Add 1 to both sides to solve for y:
y= 1 ±2√1−(x−3)2.
Therefore, the equation is solved for yin terms of xas y= 1±2√1−(x−3)2.
Question 2
Question
Solve the following system of equations:
{2x2+y2= 25
x−2y= 3
Solution
Step 1: Let’s isolate xin the second equation to express it in terms of y:
x= 3 + 2y
Step 2: Substitute xin the first equation with 3+2y:
2(3 + 2y)2+y2= 25
Step 3: Simplify the equation:
2(9 + 12y+ 4y2) + y2= 25
18 + 24y+ 8y2+y2= 25
9y2+ 24y−7 = 0
Step 4: Solve the quadratic equation 9y2+ 24y−7=0using the quadratic
formula:
y=−B±√B2−4AC
2A
y=−24 ±√242−4∗9∗(−7)
2∗9
y=−24 ±√576 + 252
18
y=−24 ±√828
18
y=−24 ±2√207
18
y=−12 ±√207
9
Step 5: Now, find the corresponding values of xby substituting the values
of yback into x= 3 + 2y:
x= 3 + 2 (−12 ±√207
9)
2
x= 3 + −24 ±2√207
9
x=−9±2√207
9
Therefore, the solution to the system of equations is:
(−9+2√207
9,−12 + √207
9)and (−9−2√207
9,−12 −√207
9)
Question 3
Question
Solve the following system of equations:
{x2−y2= 5
xy = 2
Solution
Step 1: We can start by expressing yin terms of xusing the second equation.
We have y=2
x.
Step 2: Substitute the expression for yinto the first equation and solve for
x:
x2−(2
x)2
= 5
x2−4
x2= 5
x4−5x2−4 = 0
Step 3: Let u=x2, then the equation becomes u2−5u−4 = 0.
Step 4: Solve for uby factoring:
(u−4)(u+ 1) = 0
Step 5: Setting each factor to zero gives u= 4 or u=−1.
Step 6: Substituting back u=x2: For u= 4, we have x2= 4 which implies
x= 2 or x=−2. For u=−1, the equation has no real solutions as the square
of a real number is non-negative.
Step 7: Finally, substitute the values of xback into the equation xy = 2 to
find the corresponding values of y.
Therefore, the solutions to the system of equations are (x, y) = (2,1) and
(−2,−1).
3
Question 4
Question
Find the standard form of the equation of the hyperbola with vertices at (−5,0)
and (5,0) and a focus at (8,0).
Solution
Step 1: Find the center of the hyperbola. The center of the hyperbola is the
midpoint of the line segment between the vertices, which in this case is (0,0).
Step 2: Find the distance between the center and one of the vertices. This
distance is the distance from the center to a focus divided by c, the distance
from center to a focus. We have c= 8, so the distance between the center and
a vertex is √c2−b2.
Step 3: Find b. Since the vertices are at (±5,0), we know a= 5. Using
the relationship a2=b2+c2, we can solve for b. Thus, 25 = b2+ 64 or
b=√25 −64 = 3√7.
Therefore, the equation of the hyperbola is of the form x2
a2−y2
b2= 1. Plugging
in our values of aand b, we have x2
25 −y2
63 = 1.
Question 5
Question
Solve the following system of equations:
{3x+ 2y2= 11
x−y= 3
Solution
Step 1: Start by isolating xin the second equation. This can be done by adding
yto both sides of the equation.
x−y= 3
x= 3 + y
Step 2: Substitute x= 3 + yinto the first equation and solve for y.
3(3 + y)+2y2= 11
9+3y+ 2y2= 11
2y2+ 3y−2 = 0
4
Step 3: Solve the quadratic equation 2y2+ 3y−2 = 0 by factoring or using
the quadratic formula.
(2y−1)(y+ 2) = 0
So,
y=1
2or y=−2
Step 4: Now that we have possible values for y, we can find the corresponding
values of xby substituting back into the equation x= 3 + y. For y=1
2:
x= 3 + 1
2
x=7
2
For y=−2:
x= 3 −2
x= 1
Step 5: Therefore, the solutions to the system of equations are (x, y) = (7
2,1
2)
and (x, y) = (1,−2).
Question 6
Question
Find the standard form of the equation of the ellipse with foci at (−3,2) and
(5,2) and a major axis of length 10 units.
Solution
Step 1: Find the center of the ellipse. Since the foci lie on the line y= 2, the
center of the ellipse must also lie on this line. The average of the x-coordinates
of the foci gives us the x-coordinate of the center. Therefore, the center is at
(((−3) + 5)/2,2) = (1,2).
Step 2: Find the distance between the foci. The distance between the foci is
equal to the length of the major axis, which is 10 units. The distance formula
between two points (x1, y1)and (x2, y2)is given by
d=√(x2−x1)2+ (y2−y1)2
Substitute the coordinates of the foci:
10 = √(5 −(−3))2+ (2 −2)2=√64 = 8
Step 3: Find the distance from the center to a focus. Since the distance
from the center to each focus is half the length of the major axis, the distance
is 5 units.
5
Step 4: Find the value of c. In the ellipse equation, crepresents the distance
from the center to a focus. Therefore, c= 5.
Step 5: Find the value of a. In an ellipse, the distance from the center to a
vertex along the major axis is a. The value of ain this case is 5 units.
Step 6: Find the value of b. Since aand care known, we can use the
Pythagorean theorem to find b. The relation between a,b, and cin an ellipse is
given by a2=b2+c2. Substituting the known values:
52=b2+ 52
25 = b2+ 25
b2= 0
b= 0
Step 7: Write the equation of the ellipse in standard form. The standard
form of the equation of an ellipse with center at (h, k), major axis along the
x-axis, major axis of length 2a, and minor axis of length 2bis:
(x−h)2
a2+(y−k)2
b2= 1
Substitute the values of h,k,a, and binto the equation:
(x−1)2
52+(y−2)2
0= 1
(x−1)2
25 = 1
(x−1)2= 25
x−1 = ±5
x= 6 or x=−4
So, the standard form of the ellipse equation is (x−1)2= 25.
Question 7
Question
Determine the standard form of the equation of the ellipse with foci at (-1,2)
and (-1,-2), and minor axis length of 8 units.
6
Solution
Step 1: Find the center of the ellipse. The center of the ellipse is the midpoint
between the two foci. Using the midpoint formula with coordinates (−1,2) and
(−1,−2):
(xcenter, ycenter) = (−1+(−1)
2,2+(−2)
2)
(xcenter, ycenter) = (−1,0)
Step 2: Determine the distance from the center to each focus. This distance
is half the length of the major axis. Since the major axis is twice the length of
the minor axis, the major axis length is 16 units.
Step 3: The distance from the center to each focus is the square root of the
difference of the squares of half of the major axis length and half of the minor
axis length. Using the distance formula:
c=√a2−b2=√82−42=√64 −16 = √48 = 4√3
Step 4: Determine the equation of the ellipse. The standard form of the
equation of an ellipse centered at (h, k)is:
(x−h)2
a2+(y−k)2
b2= 1
Substitute the values of a= 8 and c= 4√3into the equation to find b:
b2=a2−c2= 82−(4√3)2= 64 −48 = 16
b= 4
Therefore, the standard form of the equation of the ellipse is:
(x+ 1)2
64 +y2
16 = 1
Question 8
Question
Solve the following system of equations:
{x2+y2= 25
x−y= 3
7
Solution
Step 1: Solve the second equation for x:
x=y+ 3
Step 2: Substitute x=y+ 3 into the first equation:
(y+ 3)2+y2= 25
Step 3: Expand and simplify the equation:
y2+ 6y+9+y2= 25
2y2+ 6y−16 = 0
Step 4: Solve the quadratic equation 2y2+ 6y−16 = 0:
y=−b±√b2−4ac
2a
y=−6±√62−4(2)(−16)
2(2)
y=−6±√36 + 128
4
y=−6±√164
4
y=−6±2√41
4
y=−3
2±√41
2
Step 5: Find the corresponding values of xusing x=y+ 3:
x=−3
2±√41
2+ 3
Step 6: Therefore, the solutions to the system of equations are:
(3−√41
2,3 + √41
2)and (3 + √41
2,3−√41
2)
Question 9
Question
Find the standard form of the equation of the hyperbola with vertices at (−3,0)
and (3,0), and passing through the point (4,5).
8
Solution
Step 1: Find the center of the hyperbola.
The center of the hyperbola is the midpoint of the two vertices. The midpoint
formula is (x1+x2
2,y1+y2
2).
So the center is (−3+3
2,0+0
2) = (0,0). Hence, the center of the hyperbola is at
the origin.
Step 2: Determine the distance between the center and one of the vertices
to find a.
Since the hyperbola has vertices at (−3,0) and (3,0), we have |a|= 3.
Therefore, the equation is now x2
a2−y2
b2= 1.
Step 3: Determine busing the point on the hyperbola.
The equation will become x2
9−y2
b2= 1 with the point (4,5) on it.
Step 4: Substitute the point into the equation of the hyperbola.
Substitute x= 4 and y= 5 into the equation to solve for b:
16
9−25
b2= 1
Step 5: Solve for b2.16
9−1 = 25
b2
7
9=25
b2
b2=225
7
Step 6: Write the equation of the hyperbola in standard form.
x2
9−y2
225
7
= 1
x2
9−7y2
225 = 1
Therefore, the standard form of the equation of the hyperbola is x2
9−7y2
225 = 1.
Question 10
Question
Find the standard form equation of a hyperbola given its foci F1(−5,0) and
F2(5,0), and the distance between its vertices is 12.
9
Solution
Step 1: Find the coordinates of the vertices.
• The distance between the foci is 2a= 10, so a= 5.
• The distance between the vertices is 2a= 12, so a= 6.
• The vertices will be located at (−6,0) and (6,0).
Step 2: Find c, the distance from the center to each focus.
• Since c=√a2+b2, and we know a= 6, we still need to find b.
Step 3: Find b.
• We can use the relationship b2=c2−a2, with c= 5.
• Substituting a= 6 and c= 5,b2= 52−62.
• So, b2= 25 −36 = −11.
Step 4: Write the standard form equation.
• The standard form equation for a hyperbola is (x−h)2
a2−(y−k)2
b2= 1, with
the center at (h, k).
• Since the hyperbola is centered at the origin, the center is at (0,0).
• Thus, the equation of the hyperbola is x2
36 −y2
−11 = 1.
Question 11
Question
Determine the equation of the ellipse with foci F1(−4,0) and F2(4,0), and
passing through the point (2,3).
Solution
Step 1: Determine the center of the ellipse.
The center of the ellipse can be found by calculating the midpoint between
the two foci. The midpoint formula is given by
(h, k) = (x1+x2
2,y1+y2
2).
Using F1(−4,0) and F2(4,0), we find
(h, k) = (−4+4
2,0+0
2)= (0,0).
10
Therefore, the center of the ellipse is at the origin, (0,0).
Step 2: Determine the distance between the foci.
The distance between the foci is the major axis of the ellipse, 2a. The
distance formula is given by
2a=√(x2−x1)2+ (y2−y1)2.
Using F1(−4,0) and F2(4,0), we find
2a=√(4 −(−4))2+ (0 −0)2=√64 = 8.
Thus, the major axis of the ellipse is 8 units.
Step 3: Determine the minor axis of the ellipse.
Since the ellipse passes through the point (2,3), the distance from the center
(0,0) to this point is the minor axis, 2b. The distance formula is given by
2b=√(x−h)2+ (y−k)2.
Using (2,3) and (0,0), we find
2b=√(2 −0)2+ (3 −0)2=√13.
Therefore, the minor axis of the ellipse is √13 units.
Step 4: Write the equation of the ellipse.
The equation of an ellipse centered at the origin with major axis 2a along
the x-axis and minor axis 2b along the y-axis is
x2
a2+y2
b2= 1.
Substitute a= 4 and b=√13 into the equation to obtain
x2
16 +y2
13 = 1.
Thus, the equation of the ellipse is x2
16 +y2
13 = 1 .
Question 12
Question
Find the standard form equation of the hyperbola with vertices at (−5,0) and
(5,0) and foci at (−6,0) and (6,0).
11
Solution
Step 1: Find the center of the hyperbola by finding the midpoint between the
vertices:
Center =(−5+5
2,0+0
2)= (0,0)
Step 2: Determine the distance between the center and either vertex to find
a:
a= 5
Step 3: Determine the distance between the center and either focus to find
c:
c= 6
Step 4: Use the relationship between a,band cfor a hyperbola, where bis
the distance between the center and the transverse axis:
c2=a2+b2=⇒b=√c2−a2
b=√62−52=√36 −25 = √11
Step 5: The standard form equation of a hyperbola centered at (h, k)is:
(x−h)2
a2−(y−k)2
b2= 1
Plugging in the values, the standard form equation of the hyperbola is:
x2
25 −y2
11 = 1
Question 13
Question
Find the equation of the hyperbola with vertices at (−3,0) and (3,0), and foci
at (−5,0) and (5,0).
Solution
Step 1: Find the center of the hyperbola.
The center of the hyperbola is the midpoint of the segment connecting the
vertices, which is ((−3 + 3)/2,(0 + 0)/2), or (0,0).
Step 2: Find the distance from the center to either vertex.
The distance from the center to a vertex is the distance from (0,0) to (−3,0)
or (3,0), which is 3.
Step 3: Find the distance between the vertices, which is 2a.
Since the distance between the center and a vertex is 3, we have 2a= 3, so
a= 1.5.
12
Step 4: Find the distance between the foci, which is 2c.
Since the distance between the foci is 10, we have 2c= 10, so c= 5.
Step 5: Use the relationship a2+b2=c2to find b2.
Since a= 1.5and c= 5, we have b2=c2−a2= 52−1.52= 25 −2.25 = 22.75.
Step 6: Write the equation of the hyperbola in standard form.
The equation of a hyperbola with center at (h, k), vertices along the transverse
axis, and foci along the transverse axis is
(x−h)2/a2−(y−k)2/b2= 1
Plugging in the values, the equation becomes
(x−0)2/1.52−(y−0)2/22.75 = 1
Simplifying, we get
x2/2.25 −y2/22.75 = 1
Question 14
Question
Suppose a parabola has a vertex at (−2,3) and its focus is at (−2,5). Write an
equation in standard form for the parabola that opens downward.
Solution
Step 1: Recall the standard form of a parabola that opens downward with vertex
(h, k):
(x−h)2= 4p(y−k)
where the focus is F(h, k +p).
Step 2: We can identify h=−2and k= 3 from the given vertex. Since the
focus is at the point (−2,5), we know that k+p= 5.
Step 3: Solve for pusing the relationship k+p= 5:
3 + p= 5
p= 5 −3
p= 2
Step 4: Substitute h=−2,k= 3, and p= 2 into the standard form of the
equation:
(x+ 2)2= 4 ·2(y−3)
Step 5: Simplify the equation:
(x+ 2)2= 8(y−3)
13
Step 6: Expand the left side of the equation:
x2+ 4x+ 4 = 8y−24
Step 7: Rewrite the equation in standard form by isolating y:
8y=x2+ 4x+ 28
y=1
8x2+1
2x+ 3.5
Step 8: Therefore, the equation of the parabola that opens downward with
the given vertex and focus is:
y=1
8x2+1
2x+ 3.5
Question 15
Question
Solve the following system of equations:
{3x2−y2= 16
x+y= 5
Solution
Step 1: Solve the second equation for one variable in terms of the other. Let’s
solve for xin terms of y:
x= 5 −y
Step 2: Substitute x= 5 −yinto the first equation:
3(5 −y)2−y2= 16
Step 3: Expand and simplify the equation:
75 −30y+ 3y2−y2= 16
2y2−30y+ 59 = 0
Step 4: Solve the quadratic equation for yusing the quadratic formula:
y=−(−30) ±√(−30)2−4(2)(59)
2(2)
y=30 ±√900 −472
4
14
y=30 ±√428
4
y=30 ±2√107
4
y=15 ±√107
2
Step 5: There are two possible values for y:y= 15 + √107/2and y=
15 −√107/2.
Step 6: Substitute each value of yback into the equation x= 5 −yto find
the corresponding values of x: For y= 15 + √107/2:
x= 5 −15 + √107
2=−5 + √107
2
For y= 15 −√107/2:
x= 5 −15 −√107
2=√107 −5
2
Therefore, the solutions to the system of equations are x=−5+√107
2,y=
15 + √107/2and x=√107−5
2,y= 15 −√107/2.
Question 16
Question
Solve the system of equations:
{x2+y2= 4
x2−y2= 1
Solution
Step 1: Start by adding the two equations together to eliminate y2.
x2+y2+x2−y2= 4 + 1
Step 2: Simplify the equation.
2x2= 5
Step 3: Solve for x.
x2=5
2=⇒x=±√5
2
15
Step 4: Substitute the values of xback into the first equation to solve for y.
(√5
2)2+y2= 4
−(√5
2)2+y2= 4
Step 5: Solve for yin both equations.
{5
2+y2= 4
−5
2+y2= 4
Step 6: Simplify the equations.
{y2=3
2
y2=13
2
Step 7: Solve for yin both equations.
y=±√3
2
y=±√13
2
Therefore, the solutions to the system of equations are:
(√5
2,√3
2),(√5
2,−√3
2),(−√5
2,√13
2),(−√5
2,−√13
2)
Question 17
Question
Solve the equation of the parabola given below for y:
4x2−16x+ 3y−11 = 0
Solution
Step 1: First, we need to rearrange the equation in the standard form of the
equation of a parabola. The standard form of a vertical parabola is given by:
(x−h)2= 4p(y−k)
where the vertex of the parabola is at (h, k)and the distance from the vertex
to the focus is p.
Step 2: To rewrite the given equation in the standard form, complete the
square for the xterms:
4x2−16x+ 3y−11 = 0
16
4(x2−4x)+3y−11 = 0
4(x2−4x+4)+3y−11 = 4
4(x−2)2+ 3y−11 = 4
Step 3: Rearrange the equation to solve for y:
4(x−2)2+ 3y−11 = 4
3y=−4(x−2)2+ 15
y=−4
3(x−2)2+ 5
Step 4: Therefore, the equation of the parabola solved for yis:
y=−4
3(x−2)2+ 5
Question 18
Question
Given the equation of a hyperbola in standard form: (x−2)2
9−(y+1)2
16 = 1, find
the coordinates of the center, the vertices, the foci, and the equations of the
asymptotes.
Solution
Step 1: Identify the center of the hyperbola.
The center of the hyperbola is given by (h, k)in the standard form (x−h)2
a2−
(y−k)2
b2= 1. Therefore, in this case, the center is at (2,−1).
Step 2: Find the vertices.
The distance from the center to the vertices in the horizontal direction is a= 3,
while in the vertical direction is b= 4. Therefore, the vertices are at (h±a, k),
which gives the vertices as (2 ±3,−1), or (5,−1) and (−1,−1).
Step 3: Determine the foci.
The distance from the center to the foci is given by c=√a2+b2. Using the
values a= 3 and b= 4, we get c=√9 + 16 = √25 = 5. Hence, the foci are at
(h±c, k), which gives the coordinates as (2 ±5,−1), or (−3,−1) and (7,−1).
Step 4: Find the equations of the asymptotes.
The equations of the asymptotes for a hyperbola centered at (h, k)are given
by y−k=±b
a(x−h). Substituting the given values, we find the equations
as y+ 1 = ±4
3(x−2) and simplifying gives the equations of the asymptotes as
y=±4
3x−11
3.
Therefore, the coordinates of the center are (2,−1), the vertices are (5,−1)
and (−1,−1), the foci are (7,−1) and (−3,−1), and the equations of the asymp-
totes are y=4
3x−11
3and y=−4
3x−11
3.
17
Question 19
Question
Solve the system of equations:
{x2−y2= 9
x+y= 3
Solution
Step 1: Add the second equation to the first equation to eliminate y.
x2−y2+x+y= 9 + 3
x2+x−y2+y= 12
x(x+ 1) −y(y+ 1) = 12
Step 2: Substitute x= 3 −yinto the simplified equation.
(3 −y)(3 −y+ 1) −y(y+ 1) = 12
(3 −y)(4 −y)−y2−y= 12
12 −7y+y2= 12
y2−7y= 0
y(y−7) = 0
Step 3: Solve for yby setting each factor to zero.
y= 0 or y= 7
Step 4: Substitute y= 0 and y= 7 back into the equation x+y= 3 to find
x. For y= 0:
x+ 0 = 3
x= 3
For y= 7:
x+ 7 = 3
x=−4
Therefore, the solutions to the system of equations are (3,0) and (−4,7).
Question 20
Question
Find the standard form equation of a hyperbola with vertices at (−3,2) and
(5,2) and foci at (−7,2) and (9,2).
18
Solution
Step 1: First, let’s determine the center of the hyperbola, which lies at the
midpoint of the segment connecting the vertices. In this case, the center is
C(−3+5
2,2+2
2).
Step 2: The center of the hyperbola is C(1,2).
Step 3: Next, let’s find the distance from the center to one of the vertices.
This distance is the distance ain the standard form equation of a hyperbola.
Step 4: The distance from the center to a vertex is a=|1−(−3)|= 4.
Step 5: The distance from the center to a focus point is the distance cin the
standard form equation of a hyperbola.
Step 6: The distance from the center to a focus is c=| − 7−1|= 8.
Step 7: Recall the relationship between a,b, and cfor a hyperbola: a2+b2=
c2.
Step 8: Substitute aand cinto the equation: 42+b2= 82.
Step 9: Simplify the equation to solve for b:16 + b2= 64.
Step 10: Subtract 16 from both sides to get b2= 48.
Step 11: The standard form equation of a hyperbola is (x−h)2
a2−(y−k)2
b2= 1,
with the center at (h, k). In this case, h= 1,k= 2,a= 4, and b=√48.
Step 12: The standard form equation of the hyperbola is (x−1)2
16 −(y−2)2
48 = 1.
Question 21
Question
Solve the following system of equations:
{x2+y2= 10
x−y= 3
Solution
Let’s solve the system of equations step by step:
Step 1: Solve the second equation for xin terms of y:
x=y+ 3
Step 2: Substitute xin terms of yinto the first equation:
(y+ 3)2+y2= 10
Step 3: Expand and simplify:
y2+ 6y+9+y2= 10
2y2+ 6y−1 = 0
19
Step 4: Solve the quadratic equation for yusing the quadratic formula:
y=−6±√62−4·2·(−1)
2·2
y=−6±√36 + 8
4
y=−6±√44
4
y=−6±2√11
4
y=−3±√11
2
Step 5: Find the corresponding values of xusing x=y+3: For y=−3+√11
2,
we have x=−3+√11
2+ 3
For y=−3−√11
2, we have x=−3−√11
2+ 3
Step 6: So, the solutions to the system of equations are:
(x, y) = (−3 + √11
2+ 3,−3 + √11
2)
(x, y) = (−3−√11
2+ 3,−3−√11
2)
Question 22
Question
Find the standard form of the equation of the ellipse given the following infor-
mation: The ellipse has a major axis of length 10 units and passes through the
points (−3,2) and (7,2).
Solution
Step 1: Find the center of the ellipse. The center of the ellipse is the midpoint
of the major axis. The midpoit of the segment joining (−3,2) and (7,2) is
(−3+7
2,2+2
2) = (2,2).
Therefore, the center of the ellipse is (2,2).
Step 2: Find the endpoints of the major axis. The major axis is along the
x-axis, so the endpoints are (2 −5,2) and (2 + 5,2). This gives the endpoints
(−3,2) and (7,2), as expected.
20
Step 3: Find the equation of the ellipse. Since the major axis is along the
x-axis, the equation of the ellipse is of the form
(x−h)2
a2+(y−k)2
b2= 1,
where (h, k)is the center of the ellipse, 2ais the length of the major axis, and
2bis the length of the minor axis.
Given that the major axis has length 10 units, 2a= 10 so a= 5.
Plugging in the center (2,2) and a= 5 into the equation, we get
(x−2)2
25 +(y−2)2
b2= 1.
Step 4: Find b. Since the ellipse passes through the point (−3,2), we have
(−3−2)2
25 +(2 −2)2
b2= 1,
which simplifies to 25
25 +0
b2= 1,
and leads to 1 = 1 + 0, which is true for any b= 0.
Therefore, the equation of the ellipse is
(x−2)2
25 +(y−2)2
b2= 1 .
Question 23
Question
Let f(x) = −2x2−4x+ 3. Determine the vertex, focus, directrix, and axis of
symmetry of the parabola defined by f(x).
Solution
Step 1: To find the vertex of the parabola defined by f(x), we use the formula
h=−b
2a.
Step 1: h=−b
2a=−−4
2(−2) =−4
−4= 1.
Step 2: Substitute h= 1 into the function f(x)to find the y-coordinate of
the vertex.
Step 2: f(1) = −2(1)2−4(1) + 3 = −2−4 + 3 = −3.
So, the vertex is at (1,−3).
21
Step 3: The axis of symmetry of the parabola is the vertical line passing
through the vertex. In this case, the axis of symmetry is x= 1.
Step 4: To find the focus and directrix of the parabola, we use the formulas
p=1
4aand y=k−p, where kis the y-coordinate of the vertex.
Step 4: p=1
4(−2) =−1
8.
Substitute p=−1
8and k=−3into y=k−pto find the y-coordinate of the
focus and the equation of the directrix.
Step 4: y=−3−(−1
8)=−23
8.
Therefore, the focus is at (1,−23
8)and the directrix has the equation y=−15
8.
Question 24
Question
Find the standard form of the equation of the hyperbola with foci at (−3,0)
and (3,0) and vertices at (−4,0) and (4,0).
Solution
Step 1: Find the center of the hyperbola. The center is the midpoint of the foci,
which is ((−3 + 3)/2,0) = (0,0).
Step 2: Find a, the distance from the center to a vertex. a= 4.
Step 3: Find c, the distance from the center to a focus. c= 3.
Step 4: Use the relationship for hyperbolas: c2=a2+b2to solve for b. Plug
in the values of a= 4 and c= 3.
32= 42+b2
9 = 16 + b2
b2=−7
Since b2<0, this hyperbola is horizontal.
Step 5: The standard form of the equation of a hyperbola with a horizontal
transverse axis and center at the origin is:
x2
a2−y2
b2= 1
Therefore, the standard form of the equation of the hyperbola is:
x2
16 −y2
7= 1
22
Question 25
Question
Find the standard form of the equation of the parabola that satisfies the given
conditions: vertex at (−1,3) and focus at (−1,1).
Solution
Step 1: Recall that the standard form of the equation of a parabola with vertex
at (h, k)and focus at (h, k +p)or (h+p, k)is given by:
(x−h)2= 4p(y−k)
where pis the distance from the vertex to the focus.
Step 2: In this case, the vertex is (−1,3) and the focus is (−1,1). This
means that p= 3 −1 = 2.
Step 3: Substituting the values of h,k, and pinto the standard form equa-
tion, we have:
(x+ 1)2= 8(y−3)
Therefore, the standard form of the equation of the parabola is (x+ 1)2=
8(y−3).
23
Question 2
Question
Solve the following system of equations:
{2x2+y2= 25
x−2y= 3
Solution
Step 1: Let’s isolate xin the second equation to express it in terms of y:
x= 3 + 2y
Step 2: Substitute xin the first equation with 3+2y:
2(3 + 2y)2+y2= 25
Step 3: Simplify the equation:
2(9 + 12y+ 4y2) + y2= 25
18 + 24y+ 8y2+y2= 25
9y2+ 24y−7 = 0
Step 4: Solve the quadratic equation 9y2+ 24y−7=0using the quadratic
formula:
y=−B±√B2−4AC
2A
y=−24 ±√242−4∗9∗(−7)
2∗9
y=−24 ±√576 + 252
18
y=−24 ±√828
18
y=−24 ±2√207
18
y=−12 ±√207
9
Step 5: Now, find the corresponding values of xby substituting the values
of yback into x= 3 + 2y:
x= 3 + 2 (−12 ±√207
9)
2
x= 3 + −24 ±2√207
9
x=−9±2√207
9
Therefore, the solution to the system of equations is:
(−9+2√207
9,−12 + √207
9)and (−9−2√207
9,−12 −√207
9)
Question 3
Question
Solve the following system of equations:
{x2−y2= 5
xy = 2
Solution
Step 1: We can start by expressing yin terms of xusing the second equation.
We have y=2
x.
Step 2: Substitute the expression for yinto the first equation and solve for
x:
x2−(2
x)2
= 5
x2−4
x2= 5
x4−5x2−4 = 0
Step 3: Let u=x2, then the equation becomes u2−5u−4 = 0.
Step 4: Solve for uby factoring:
(u−4)(u+ 1) = 0
Step 5: Setting each factor to zero gives u= 4 or u=−1.
Step 6: Substituting back u=x2: For u= 4, we have x2= 4 which implies
x= 2 or x=−2. For u=−1, the equation has no real solutions as the square
of a real number is non-negative.
Step 7: Finally, substitute the values of xback into the equation xy = 2 to
find the corresponding values of y.
Therefore, the solutions to the system of equations are (x, y) = (2,1) and
(−2,−1).
3
Question 4
Question
Find the standard form of the equation of the hyperbola with vertices at (−5,0)
and (5,0) and a focus at (8,0).
Solution
Step 1: Find the center of the hyperbola. The center of the hyperbola is the
midpoint of the line segment between the vertices, which in this case is (0,0).
Step 2: Find the distance between the center and one of the vertices. This
distance is the distance from the center to a focus divided by c, the distance
from center to a focus. We have c= 8, so the distance between the center and
a vertex is √c2−b2.
Step 3: Find b. Since the vertices are at (±5,0), we know a= 5. Using
the relationship a2=b2+c2, we can solve for b. Thus, 25 = b2+ 64 or
b=√25 −64 = 3√7.
Therefore, the equation of the hyperbola is of the form x2
a2−y2
b2= 1. Plugging
in our values of aand b, we have x2
25 −y2
63 = 1.
Question 5
Question
Solve the following system of equations:
{3x+ 2y2= 11
x−y= 3
Solution
Step 1: Start by isolating xin the second equation. This can be done by adding
yto both sides of the equation.
x−y= 3
x= 3 + y
Step 2: Substitute x= 3 + yinto the first equation and solve for y.
3(3 + y)+2y2= 11
9+3y+ 2y2= 11
2y2+ 3y−2 = 0
4
Step 3: Solve the quadratic equation 2y2+ 3y−2 = 0 by factoring or using
the quadratic formula.
(2y−1)(y+ 2) = 0
So,
y=1
2or y=−2
Step 4: Now that we have possible values for y, we can find the corresponding
values of xby substituting back into the equation x= 3 + y. For y=1
2:
x= 3 + 1
2
x=7
2
For y=−2:
x= 3 −2
x= 1
Step 5: Therefore, the solutions to the system of equations are (x, y) = (7
2,1
2)
and (x, y) = (1,−2).
Question 6
Question
Find the standard form of the equation of the ellipse with foci at (−3,2) and
(5,2) and a major axis of length 10 units.
Solution
Step 1: Find the center of the ellipse. Since the foci lie on the line y= 2, the
center of the ellipse must also lie on this line. The average of the x-coordinates
of the foci gives us the x-coordinate of the center. Therefore, the center is at
(((−3) + 5)/2,2) = (1,2).
Step 2: Find the distance between the foci. The distance between the foci is
equal to the length of the major axis, which is 10 units. The distance formula
between two points (x1, y1)and (x2, y2)is given by
d=√(x2−x1)2+ (y2−y1)2
Substitute the coordinates of the foci:
10 = √(5 −(−3))2+ (2 −2)2=√64 = 8
Step 3: Find the distance from the center to a focus. Since the distance
from the center to each focus is half the length of the major axis, the distance
is 5 units.
5
Step 4: Find the value of c. In the ellipse equation, crepresents the distance
from the center to a focus. Therefore, c= 5.
Step 5: Find the value of a. In an ellipse, the distance from the center to a
vertex along the major axis is a. The value of ain this case is 5 units.
Step 6: Find the value of b. Since aand care known, we can use the
Pythagorean theorem to find b. The relation between a,b, and cin an ellipse is
given by a2=b2+c2. Substituting the known values:
52=b2+ 52
25 = b2+ 25
b2= 0
b= 0
Step 7: Write the equation of the ellipse in standard form. The standard
form of the equation of an ellipse with center at (h, k), major axis along the
x-axis, major axis of length 2a, and minor axis of length 2bis:
(x−h)2
a2+(y−k)2
b2= 1
Substitute the values of h,k,a, and binto the equation:
(x−1)2
52+(y−2)2
0= 1
(x−1)2
25 = 1
(x−1)2= 25
x−1 = ±5
x= 6 or x=−4
So, the standard form of the ellipse equation is (x−1)2= 25.
Question 7
Question
Determine the standard form of the equation of the ellipse with foci at (-1,2)
and (-1,-2), and minor axis length of 8 units.
6
Solution
Step 1: Find the center of the ellipse. The center of the ellipse is the midpoint
between the two foci. Using the midpoint formula with coordinates (−1,2) and
(−1,−2):
(xcenter, ycenter) = (−1+(−1)
2,2+(−2)
2)
(xcenter, ycenter) = (−1,0)
Step 2: Determine the distance from the center to each focus. This distance
is half the length of the major axis. Since the major axis is twice the length of
the minor axis, the major axis length is 16 units.
Step 3: The distance from the center to each focus is the square root of the
difference of the squares of half of the major axis length and half of the minor
axis length. Using the distance formula:
c=√a2−b2=√82−42=√64 −16 = √48 = 4√3
Step 4: Determine the equation of the ellipse. The standard form of the
equation of an ellipse centered at (h, k)is:
(x−h)2
a2+(y−k)2
b2= 1
Substitute the values of a= 8 and c= 4√3into the equation to find b:
b2=a2−c2= 82−(4√3)2= 64 −48 = 16
b= 4
Therefore, the standard form of the equation of the ellipse is:
(x+ 1)2
64 +y2
16 = 1
Question 8
Question
Solve the following system of equations:
{x2+y2= 25
x−y= 3
7
Solution
Step 1: Solve the second equation for x:
x=y+ 3
Step 2: Substitute x=y+ 3 into the first equation:
(y+ 3)2+y2= 25
Step 3: Expand and simplify the equation:
y2+ 6y+9+y2= 25
2y2+ 6y−16 = 0
Step 4: Solve the quadratic equation 2y2+ 6y−16 = 0:
y=−b±√b2−4ac
2a
y=−6±√62−4(2)(−16)
2(2)
y=−6±√36 + 128
4
y=−6±√164
4
y=−6±2√41
4
y=−3
2±√41
2
Step 5: Find the corresponding values of xusing x=y+ 3:
x=−3
2±√41
2+ 3
Step 6: Therefore, the solutions to the system of equations are:
(3−√41
2,3 + √41
2)and (3 + √41
2,3−√41
2)
Question 9
Question
Find the standard form of the equation of the hyperbola with vertices at (−3,0)
and (3,0), and passing through the point (4,5).
8
Solution
Step 1: Find the center of the hyperbola.
The center of the hyperbola is the midpoint of the two vertices. The midpoint
formula is (x1+x2
2,y1+y2
2).
So the center is (−3+3
2,0+0
2) = (0,0). Hence, the center of the hyperbola is at
the origin.
Step 2: Determine the distance between the center and one of the vertices
to find a.
Since the hyperbola has vertices at (−3,0) and (3,0), we have |a|= 3.
Therefore, the equation is now x2
a2−y2
b2= 1.
Step 3: Determine busing the point on the hyperbola.
The equation will become x2
9−y2
b2= 1 with the point (4,5) on it.
Step 4: Substitute the point into the equation of the hyperbola.
Substitute x= 4 and y= 5 into the equation to solve for b:
16
9−25
b2= 1
Step 5: Solve for b2.16
9−1 = 25
b2
7
9=25
b2
b2=225
7
Step 6: Write the equation of the hyperbola in standard form.
x2
9−y2
225
7
= 1
x2
9−7y2
225 = 1
Therefore, the standard form of the equation of the hyperbola is x2
9−7y2
225 = 1.
Question 10
Question
Find the standard form equation of a hyperbola given its foci F1(−5,0) and
F2(5,0), and the distance between its vertices is 12.
9
Solution
Step 1: Find the coordinates of the vertices.
• The distance between the foci is 2a= 10, so a= 5.
• The distance between the vertices is 2a= 12, so a= 6.
• The vertices will be located at (−6,0) and (6,0).
Step 2: Find c, the distance from the center to each focus.
• Since c=√a2+b2, and we know a= 6, we still need to find b.
Step 3: Find b.
• We can use the relationship b2=c2−a2, with c= 5.
• Substituting a= 6 and c= 5,b2= 52−62.
• So, b2= 25 −36 = −11.
Step 4: Write the standard form equation.
• The standard form equation for a hyperbola is (x−h)2
a2−(y−k)2
b2= 1, with
the center at (h, k).
• Since the hyperbola is centered at the origin, the center is at (0,0).
• Thus, the equation of the hyperbola is x2
36 −y2
−11 = 1.
Question 11
Question
Determine the equation of the ellipse with foci F1(−4,0) and F2(4,0), and
passing through the point (2,3).
Solution
Step 1: Determine the center of the ellipse.
The center of the ellipse can be found by calculating the midpoint between
the two foci. The midpoint formula is given by
(h, k) = (x1+x2
2,y1+y2
2).
Using F1(−4,0) and F2(4,0), we find
(h, k) = (−4+4
2,0+0
2)= (0,0).
10
Therefore, the center of the ellipse is at the origin, (0,0).
Step 2: Determine the distance between the foci.
The distance between the foci is the major axis of the ellipse, 2a. The
distance formula is given by
2a=√(x2−x1)2+ (y2−y1)2.
Using F1(−4,0) and F2(4,0), we find
2a=√(4 −(−4))2+ (0 −0)2=√64 = 8.
Thus, the major axis of the ellipse is 8 units.
Step 3: Determine the minor axis of the ellipse.
Since the ellipse passes through the point (2,3), the distance from the center
(0,0) to this point is the minor axis, 2b. The distance formula is given by
2b=√(x−h)2+ (y−k)2.
Using (2,3) and (0,0), we find
2b=√(2 −0)2+ (3 −0)2=√13.
Therefore, the minor axis of the ellipse is √13 units.
Step 4: Write the equation of the ellipse.
The equation of an ellipse centered at the origin with major axis 2a along
the x-axis and minor axis 2b along the y-axis is
x2
a2+y2
b2= 1.
Substitute a= 4 and b=√13 into the equation to obtain
x2
16 +y2
13 = 1.
Thus, the equation of the ellipse is x2
16 +y2
13 = 1 .
Question 12
Question
Find the standard form equation of the hyperbola with vertices at (−5,0) and
(5,0) and foci at (−6,0) and (6,0).
11
Solution
Step 1: Find the center of the hyperbola by finding the midpoint between the
vertices:
Center =(−5+5
2,0+0
2)= (0,0)
Step 2: Determine the distance between the center and either vertex to find
a:
a= 5
Step 3: Determine the distance between the center and either focus to find
c:
c= 6
Step 4: Use the relationship between a,band cfor a hyperbola, where bis
the distance between the center and the transverse axis:
c2=a2+b2=⇒b=√c2−a2
b=√62−52=√36 −25 = √11
Step 5: The standard form equation of a hyperbola centered at (h, k)is:
(x−h)2
a2−(y−k)2
b2= 1
Plugging in the values, the standard form equation of the hyperbola is:
x2
25 −y2
11 = 1
Question 13
Question
Find the equation of the hyperbola with vertices at (−3,0) and (3,0), and foci
at (−5,0) and (5,0).
Solution
Step 1: Find the center of the hyperbola.
The center of the hyperbola is the midpoint of the segment connecting the
vertices, which is ((−3 + 3)/2,(0 + 0)/2), or (0,0).
Step 2: Find the distance from the center to either vertex.
The distance from the center to a vertex is the distance from (0,0) to (−3,0)
or (3,0), which is 3.
Step 3: Find the distance between the vertices, which is 2a.
Since the distance between the center and a vertex is 3, we have 2a= 3, so
a= 1.5.
12
Step 4: Find the distance between the foci, which is 2c.
Since the distance between the foci is 10, we have 2c= 10, so c= 5.
Step 5: Use the relationship a2+b2=c2to find b2.
Since a= 1.5and c= 5, we have b2=c2−a2= 52−1.52= 25 −2.25 = 22.75.
Step 6: Write the equation of the hyperbola in standard form.
The equation of a hyperbola with center at (h, k), vertices along the transverse
axis, and foci along the transverse axis is
(x−h)2/a2−(y−k)2/b2= 1
Plugging in the values, the equation becomes
(x−0)2/1.52−(y−0)2/22.75 = 1
Simplifying, we get
x2/2.25 −y2/22.75 = 1
Question 14
Question
Suppose a parabola has a vertex at (−2,3) and its focus is at (−2,5). Write an
equation in standard form for the parabola that opens downward.
Solution
Step 1: Recall the standard form of a parabola that opens downward with vertex
(h, k):
(x−h)2= 4p(y−k)
where the focus is F(h, k +p).
Step 2: We can identify h=−2and k= 3 from the given vertex. Since the
focus is at the point (−2,5), we know that k+p= 5.
Step 3: Solve for pusing the relationship k+p= 5:
3 + p= 5
p= 5 −3
p= 2
Step 4: Substitute h=−2,k= 3, and p= 2 into the standard form of the
equation:
(x+ 2)2= 4 ·2(y−3)
Step 5: Simplify the equation:
(x+ 2)2= 8(y−3)
13
Step 6: Expand the left side of the equation:
x2+ 4x+ 4 = 8y−24
Step 7: Rewrite the equation in standard form by isolating y:
8y=x2+ 4x+ 28
y=1
8x2+1
2x+ 3.5
Step 8: Therefore, the equation of the parabola that opens downward with
the given vertex and focus is:
y=1
8x2+1
2x+ 3.5
Question 15
Question
Solve the following system of equations:
{3x2−y2= 16
x+y= 5
Solution
Step 1: Solve the second equation for one variable in terms of the other. Let’s
solve for xin terms of y:
x= 5 −y
Step 2: Substitute x= 5 −yinto the first equation:
3(5 −y)2−y2= 16
Step 3: Expand and simplify the equation:
75 −30y+ 3y2−y2= 16
2y2−30y+ 59 = 0
Step 4: Solve the quadratic equation for yusing the quadratic formula:
y=−(−30) ±√(−30)2−4(2)(59)
2(2)
y=30 ±√900 −472
4
14
y=30 ±√428
4
y=30 ±2√107
4
y=15 ±√107
2
Step 5: There are two possible values for y:y= 15 + √107/2and y=
15 −√107/2.
Step 6: Substitute each value of yback into the equation x= 5 −yto find
the corresponding values of x: For y= 15 + √107/2:
x= 5 −15 + √107
2=−5 + √107
2
For y= 15 −√107/2:
x= 5 −15 −√107
2=√107 −5
2
Therefore, the solutions to the system of equations are x=−5+√107
2,y=
15 + √107/2and x=√107−5
2,y= 15 −√107/2.
Question 16
Question
Solve the system of equations:
{x2+y2= 4
x2−y2= 1
Solution
Step 1: Start by adding the two equations together to eliminate y2.
x2+y2+x2−y2= 4 + 1
Step 2: Simplify the equation.
2x2= 5
Step 3: Solve for x.
x2=5
2=⇒x=±√5
2
15
Step 4: Substitute the values of xback into the first equation to solve for y.
(√5
2)2+y2= 4
−(√5
2)2+y2= 4
Step 5: Solve for yin both equations.
{5
2+y2= 4
−5
2+y2= 4
Step 6: Simplify the equations.
{y2=3
2
y2=13
2
Step 7: Solve for yin both equations.
y=±√3
2
y=±√13
2
Therefore, the solutions to the system of equations are:
(√5
2,√3
2),(√5
2,−√3
2),(−√5
2,√13
2),(−√5
2,−√13
2)
Question 17
Question
Solve the equation of the parabola given below for y:
4x2−16x+ 3y−11 = 0
Solution
Step 1: First, we need to rearrange the equation in the standard form of the
equation of a parabola. The standard form of a vertical parabola is given by:
(x−h)2= 4p(y−k)
where the vertex of the parabola is at (h, k)and the distance from the vertex
to the focus is p.
Step 2: To rewrite the given equation in the standard form, complete the
square for the xterms:
4x2−16x+ 3y−11 = 0
16
4(x2−4x)+3y−11 = 0
4(x2−4x+4)+3y−11 = 4
4(x−2)2+ 3y−11 = 4
Step 3: Rearrange the equation to solve for y:
4(x−2)2+ 3y−11 = 4
3y=−4(x−2)2+ 15
y=−4
3(x−2)2+ 5
Step 4: Therefore, the equation of the parabola solved for yis:
y=−4
3(x−2)2+ 5
Question 18
Question
Given the equation of a hyperbola in standard form: (x−2)2
9−(y+1)2
16 = 1, find
the coordinates of the center, the vertices, the foci, and the equations of the
asymptotes.
Solution
Step 1: Identify the center of the hyperbola.
The center of the hyperbola is given by (h, k)in the standard form (x−h)2
a2−
(y−k)2
b2= 1. Therefore, in this case, the center is at (2,−1).
Step 2: Find the vertices.
The distance from the center to the vertices in the horizontal direction is a= 3,
while in the vertical direction is b= 4. Therefore, the vertices are at (h±a, k),
which gives the vertices as (2 ±3,−1), or (5,−1) and (−1,−1).
Step 3: Determine the foci.
The distance from the center to the foci is given by c=√a2+b2. Using the
values a= 3 and b= 4, we get c=√9 + 16 = √25 = 5. Hence, the foci are at
(h±c, k), which gives the coordinates as (2 ±5,−1), or (−3,−1) and (7,−1).
Step 4: Find the equations of the asymptotes.
The equations of the asymptotes for a hyperbola centered at (h, k)are given
by y−k=±b
a(x−h). Substituting the given values, we find the equations
as y+ 1 = ±4
3(x−2) and simplifying gives the equations of the asymptotes as
y=±4
3x−11
3.
Therefore, the coordinates of the center are (2,−1), the vertices are (5,−1)
and (−1,−1), the foci are (7,−1) and (−3,−1), and the equations of the asymp-
totes are y=4
3x−11
3and y=−4
3x−11
3.
17
Question 19
Question
Solve the system of equations:
{x2−y2= 9
x+y= 3
Solution
Step 1: Add the second equation to the first equation to eliminate y.
x2−y2+x+y= 9 + 3
x2+x−y2+y= 12
x(x+ 1) −y(y+ 1) = 12
Step 2: Substitute x= 3 −yinto the simplified equation.
(3 −y)(3 −y+ 1) −y(y+ 1) = 12
(3 −y)(4 −y)−y2−y= 12
12 −7y+y2= 12
y2−7y= 0
y(y−7) = 0
Step 3: Solve for yby setting each factor to zero.
y= 0 or y= 7
Step 4: Substitute y= 0 and y= 7 back into the equation x+y= 3 to find
x. For y= 0:
x+ 0 = 3
x= 3
For y= 7:
x+ 7 = 3
x=−4
Therefore, the solutions to the system of equations are (3,0) and (−4,7).
Question 20
Question
Find the standard form equation of a hyperbola with vertices at (−3,2) and
(5,2) and foci at (−7,2) and (9,2).
18
Solution
Step 1: First, let’s determine the center of the hyperbola, which lies at the
midpoint of the segment connecting the vertices. In this case, the center is
C(−3+5
2,2+2
2).
Step 2: The center of the hyperbola is C(1,2).
Step 3: Next, let’s find the distance from the center to one of the vertices.
This distance is the distance ain the standard form equation of a hyperbola.
Step 4: The distance from the center to a vertex is a=|1−(−3)|= 4.
Step 5: The distance from the center to a focus point is the distance cin the
standard form equation of a hyperbola.
Step 6: The distance from the center to a focus is c=| − 7−1|= 8.
Step 7: Recall the relationship between a,b, and cfor a hyperbola: a2+b2=
c2.
Step 8: Substitute aand cinto the equation: 42+b2= 82.
Step 9: Simplify the equation to solve for b:16 + b2= 64.
Step 10: Subtract 16 from both sides to get b2= 48.
Step 11: The standard form equation of a hyperbola is (x−h)2
a2−(y−k)2
b2= 1,
with the center at (h, k). In this case, h= 1,k= 2,a= 4, and b=√48.
Step 12: The standard form equation of the hyperbola is (x−1)2
16 −(y−2)2
48 = 1.
Question 21
Question
Solve the following system of equations:
{x2+y2= 10
x−y= 3
Solution
Let’s solve the system of equations step by step:
Step 1: Solve the second equation for xin terms of y:
x=y+ 3
Step 2: Substitute xin terms of yinto the first equation:
(y+ 3)2+y2= 10
Step 3: Expand and simplify:
y2+ 6y+9+y2= 10
2y2+ 6y−1 = 0
19
Step 4: Solve the quadratic equation for yusing the quadratic formula:
y=−6±√62−4·2·(−1)
2·2
y=−6±√36 + 8
4
y=−6±√44
4
y=−6±2√11
4
y=−3±√11
2
Step 5: Find the corresponding values of xusing x=y+3: For y=−3+√11
2,
we have x=−3+√11
2+ 3
For y=−3−√11
2, we have x=−3−√11
2+ 3
Step 6: So, the solutions to the system of equations are:
(x, y) = (−3 + √11
2+ 3,−3 + √11
2)
(x, y) = (−3−√11
2+ 3,−3−√11
2)
Question 22
Question
Find the standard form of the equation of the ellipse given the following infor-
mation: The ellipse has a major axis of length 10 units and passes through the
points (−3,2) and (7,2).
Solution
Step 1: Find the center of the ellipse. The center of the ellipse is the midpoint
of the major axis. The midpoit of the segment joining (−3,2) and (7,2) is
(−3+7
2,2+2
2) = (2,2).
Therefore, the center of the ellipse is (2,2).
Step 2: Find the endpoints of the major axis. The major axis is along the
x-axis, so the endpoints are (2 −5,2) and (2 + 5,2). This gives the endpoints
(−3,2) and (7,2), as expected.
20
Step 3: Find the equation of the ellipse. Since the major axis is along the
x-axis, the equation of the ellipse is of the form
(x−h)2
a2+(y−k)2
b2= 1,
where (h, k)is the center of the ellipse, 2ais the length of the major axis, and
2bis the length of the minor axis.
Given that the major axis has length 10 units, 2a= 10 so a= 5.
Plugging in the center (2,2) and a= 5 into the equation, we get
(x−2)2
25 +(y−2)2
b2= 1.
Step 4: Find b. Since the ellipse passes through the point (−3,2), we have
(−3−2)2
25 +(2 −2)2
b2= 1,
which simplifies to 25
25 +0
b2= 1,
and leads to 1 = 1 + 0, which is true for any b= 0.
Therefore, the equation of the ellipse is
(x−2)2
25 +(y−2)2
b2= 1 .
Question 23
Question
Let f(x) = −2x2−4x+ 3. Determine the vertex, focus, directrix, and axis of
symmetry of the parabola defined by f(x).
Solution
Step 1: To find the vertex of the parabola defined by f(x), we use the formula
h=−b
2a.
Step 1: h=−b
2a=−−4
2(−2) =−4
−4= 1.
Step 2: Substitute h= 1 into the function f(x)to find the y-coordinate of
the vertex.
Step 2: f(1) = −2(1)2−4(1) + 3 = −2−4 + 3 = −3.
So, the vertex is at (1,−3).
21
Step 3: The axis of symmetry of the parabola is the vertical line passing
through the vertex. In this case, the axis of symmetry is x= 1.
Step 4: To find the focus and directrix of the parabola, we use the formulas
p=1
4aand y=k−p, where kis the y-coordinate of the vertex.
Step 4: p=1
4(−2) =−1
8.
Substitute p=−1
8and k=−3into y=k−pto find the y-coordinate of the
focus and the equation of the directrix.
Step 4: y=−3−(−1
8)=−23
8.
Therefore, the focus is at (1,−23
8)and the directrix has the equation y=−15
8.
Question 24
Question
Find the standard form of the equation of the hyperbola with foci at (−3,0)
and (3,0) and vertices at (−4,0) and (4,0).
Solution
Step 1: Find the center of the hyperbola. The center is the midpoint of the foci,
which is ((−3 + 3)/2,0) = (0,0).
Step 2: Find a, the distance from the center to a vertex. a= 4.
Step 3: Find c, the distance from the center to a focus. c= 3.
Step 4: Use the relationship for hyperbolas: c2=a2+b2to solve for b. Plug
in the values of a= 4 and c= 3.
32= 42+b2
9 = 16 + b2
b2=−7
Since b2<0, this hyperbola is horizontal.
Step 5: The standard form of the equation of a hyperbola with a horizontal
transverse axis and center at the origin is:
x2
a2−y2
b2= 1
Therefore, the standard form of the equation of the hyperbola is:
x2
16 −y2
7= 1
22
Question 25
Question
Find the standard form of the equation of the parabola that satisfies the given
conditions: vertex at (−1,3) and focus at (−1,1).
Solution
Step 1: Recall that the standard form of the equation of a parabola with vertex
at (h, k)and focus at (h, k +p)or (h+p, k)is given by:
(x−h)2= 4p(y−k)
where pis the distance from the vertex to the focus.
Step 2: In this case, the vertex is (−1,3) and the focus is (−1,1). This
means that p= 3 −1 = 2.
Step 3: Substituting the values of h,k, and pinto the standard form equa-
tion, we have:
(x+ 1)2= 8(y−3)
Therefore, the standard form of the equation of the parabola is (x+ 1)2=
8(y−3).
23
Question 2
Question
Solve the following system of equations:
{2x2+y2= 25
x−2y= 3
Solution
Step 1: Let’s isolate xin the second equation to express it in terms of y:
x= 3 + 2y
Step 2: Substitute xin the first equation with 3+2y:
2(3 + 2y)2+y2= 25
Step 3: Simplify the equation:
2(9 + 12y+ 4y2) + y2= 25
18 + 24y+ 8y2+y2= 25
9y2+ 24y−7 = 0
Step 4: Solve the quadratic equation 9y2+ 24y−7=0using the quadratic
formula:
y=−B±√B2−4AC
2A
y=−24 ±√242−4∗9∗(−7)
2∗9
y=−24 ±√576 + 252
18
y=−24 ±√828
18
y=−24 ±2√207
18
y=−12 ±√207
9
Step 5: Now, find the corresponding values of xby substituting the values
of yback into x= 3 + 2y:
x= 3 + 2 (−12 ±√207
9)
2
x= 3 + −24 ±2√207
9
x=−9±2√207
9
Therefore, the solution to the system of equations is:
(−9+2√207
9,−12 + √207
9)and (−9−2√207
9,−12 −√207
9)
Question 3
Question
Solve the following system of equations:
{x2−y2= 5
xy = 2
Solution
Step 1: We can start by expressing yin terms of xusing the second equation.
We have y=2
x.
Step 2: Substitute the expression for yinto the first equation and solve for
x:
x2−(2
x)2
= 5
x2−4
x2= 5
x4−5x2−4 = 0
Step 3: Let u=x2, then the equation becomes u2−5u−4 = 0.
Step 4: Solve for uby factoring:
(u−4)(u+ 1) = 0
Step 5: Setting each factor to zero gives u= 4 or u=−1.
Step 6: Substituting back u=x2: For u= 4, we have x2= 4 which implies
x= 2 or x=−2. For u=−1, the equation has no real solutions as the square
of a real number is non-negative.
Step 7: Finally, substitute the values of xback into the equation xy = 2 to
find the corresponding values of y.
Therefore, the solutions to the system of equations are (x, y) = (2,1) and
(−2,−1).
3
Question 4
Question
Find the standard form of the equation of the hyperbola with vertices at (−5,0)
and (5,0) and a focus at (8,0).
Solution
Step 1: Find the center of the hyperbola. The center of the hyperbola is the
midpoint of the line segment between the vertices, which in this case is (0,0).
Step 2: Find the distance between the center and one of the vertices. This
distance is the distance from the center to a focus divided by c, the distance
from center to a focus. We have c= 8, so the distance between the center and
a vertex is √c2−b2.
Step 3: Find b. Since the vertices are at (±5,0), we know a= 5. Using
the relationship a2=b2+c2, we can solve for b. Thus, 25 = b2+ 64 or
b=√25 −64 = 3√7.
Therefore, the equation of the hyperbola is of the form x2
a2−y2
b2= 1. Plugging
in our values of aand b, we have x2
25 −y2
63 = 1.
Question 5
Question
Solve the following system of equations:
{3x+ 2y2= 11
x−y= 3
Solution
Step 1: Start by isolating xin the second equation. This can be done by adding
yto both sides of the equation.
x−y= 3
x= 3 + y
Step 2: Substitute x= 3 + yinto the first equation and solve for y.
3(3 + y)+2y2= 11
9+3y+ 2y2= 11
2y2+ 3y−2 = 0
4
Step 3: Solve the quadratic equation 2y2+ 3y−2 = 0 by factoring or using
the quadratic formula.
(2y−1)(y+ 2) = 0
So,
y=1
2or y=−2
Step 4: Now that we have possible values for y, we can find the corresponding
values of xby substituting back into the equation x= 3 + y. For y=1
2:
x= 3 + 1
2
x=7
2
For y=−2:
x= 3 −2
x= 1
Step 5: Therefore, the solutions to the system of equations are (x, y) = (7
2,1
2)
and (x, y) = (1,−2).
Question 6
Question
Find the standard form of the equation of the ellipse with foci at (−3,2) and
(5,2) and a major axis of length 10 units.
Solution
Step 1: Find the center of the ellipse. Since the foci lie on the line y= 2, the
center of the ellipse must also lie on this line. The average of the x-coordinates
of the foci gives us the x-coordinate of the center. Therefore, the center is at
(((−3) + 5)/2,2) = (1,2).
Step 2: Find the distance between the foci. The distance between the foci is
equal to the length of the major axis, which is 10 units. The distance formula
between two points (x1, y1)and (x2, y2)is given by
d=√(x2−x1)2+ (y2−y1)2
Substitute the coordinates of the foci:
10 = √(5 −(−3))2+ (2 −2)2=√64 = 8
Step 3: Find the distance from the center to a focus. Since the distance
from the center to each focus is half the length of the major axis, the distance
is 5 units.
5
Step 4: Find the value of c. In the ellipse equation, crepresents the distance
from the center to a focus. Therefore, c= 5.
Step 5: Find the value of a. In an ellipse, the distance from the center to a
vertex along the major axis is a. The value of ain this case is 5 units.
Step 6: Find the value of b. Since aand care known, we can use the
Pythagorean theorem to find b. The relation between a,b, and cin an ellipse is
given by a2=b2+c2. Substituting the known values:
52=b2+ 52
25 = b2+ 25
b2= 0
b= 0
Step 7: Write the equation of the ellipse in standard form. The standard
form of the equation of an ellipse with center at (h, k), major axis along the
x-axis, major axis of length 2a, and minor axis of length 2bis:
(x−h)2
a2+(y−k)2
b2= 1
Substitute the values of h,k,a, and binto the equation:
(x−1)2
52+(y−2)2
0= 1
(x−1)2
25 = 1
(x−1)2= 25
x−1 = ±5
x= 6 or x=−4
So, the standard form of the ellipse equation is (x−1)2= 25.
Question 7
Question
Determine the standard form of the equation of the ellipse with foci at (-1,2)
and (-1,-2), and minor axis length of 8 units.
6
Solution
Step 1: Find the center of the ellipse. The center of the ellipse is the midpoint
between the two foci. Using the midpoint formula with coordinates (−1,2) and
(−1,−2):
(xcenter, ycenter) = (−1+(−1)
2,2+(−2)
2)
(xcenter, ycenter) = (−1,0)
Step 2: Determine the distance from the center to each focus. This distance
is half the length of the major axis. Since the major axis is twice the length of
the minor axis, the major axis length is 16 units.
Step 3: The distance from the center to each focus is the square root of the
difference of the squares of half of the major axis length and half of the minor
axis length. Using the distance formula:
c=√a2−b2=√82−42=√64 −16 = √48 = 4√3
Step 4: Determine the equation of the ellipse. The standard form of the
equation of an ellipse centered at (h, k)is:
(x−h)2
a2+(y−k)2
b2= 1
Substitute the values of a= 8 and c= 4√3into the equation to find b:
b2=a2−c2= 82−(4√3)2= 64 −48 = 16
b= 4
Therefore, the standard form of the equation of the ellipse is:
(x+ 1)2
64 +y2
16 = 1
Question 8
Question
Solve the following system of equations:
{x2+y2= 25
x−y= 3
7
Solution
Step 1: Solve the second equation for x:
x=y+ 3
Step 2: Substitute x=y+ 3 into the first equation:
(y+ 3)2+y2= 25
Step 3: Expand and simplify the equation:
y2+ 6y+9+y2= 25
2y2+ 6y−16 = 0
Step 4: Solve the quadratic equation 2y2+ 6y−16 = 0:
y=−b±√b2−4ac
2a
y=−6±√62−4(2)(−16)
2(2)
y=−6±√36 + 128
4
y=−6±√164
4
y=−6±2√41
4
y=−3
2±√41
2
Step 5: Find the corresponding values of xusing x=y+ 3:
x=−3
2±√41
2+ 3
Step 6: Therefore, the solutions to the system of equations are:
(3−√41
2,3 + √41
2)and (3 + √41
2,3−√41
2)
Question 9
Question
Find the standard form of the equation of the hyperbola with vertices at (−3,0)
and (3,0), and passing through the point (4,5).
8
Solution
Step 1: Find the center of the hyperbola.
The center of the hyperbola is the midpoint of the two vertices. The midpoint
formula is (x1+x2
2,y1+y2
2).
So the center is (−3+3
2,0+0
2) = (0,0). Hence, the center of the hyperbola is at
the origin.
Step 2: Determine the distance between the center and one of the vertices
to find a.
Since the hyperbola has vertices at (−3,0) and (3,0), we have |a|= 3.
Therefore, the equation is now x2
a2−y2
b2= 1.
Step 3: Determine busing the point on the hyperbola.
The equation will become x2
9−y2
b2= 1 with the point (4,5) on it.
Step 4: Substitute the point into the equation of the hyperbola.
Substitute x= 4 and y= 5 into the equation to solve for b:
16
9−25
b2= 1
Step 5: Solve for b2.16
9−1 = 25
b2
7
9=25
b2
b2=225
7
Step 6: Write the equation of the hyperbola in standard form.
x2
9−y2
225
7
= 1
x2
9−7y2
225 = 1
Therefore, the standard form of the equation of the hyperbola is x2
9−7y2
225 = 1.
Question 10
Question
Find the standard form equation of a hyperbola given its foci F1(−5,0) and
F2(5,0), and the distance between its vertices is 12.
9
Solution
Step 1: Find the coordinates of the vertices.
• The distance between the foci is 2a= 10, so a= 5.
• The distance between the vertices is 2a= 12, so a= 6.
• The vertices will be located at (−6,0) and (6,0).
Step 2: Find c, the distance from the center to each focus.
• Since c=√a2+b2, and we know a= 6, we still need to find b.
Step 3: Find b.
• We can use the relationship b2=c2−a2, with c= 5.
• Substituting a= 6 and c= 5,b2= 52−62.
• So, b2= 25 −36 = −11.
Step 4: Write the standard form equation.
• The standard form equation for a hyperbola is (x−h)2
a2−(y−k)2
b2= 1, with
the center at (h, k).
• Since the hyperbola is centered at the origin, the center is at (0,0).
• Thus, the equation of the hyperbola is x2
36 −y2
−11 = 1.
Question 11
Question
Determine the equation of the ellipse with foci F1(−4,0) and F2(4,0), and
passing through the point (2,3).
Solution
Step 1: Determine the center of the ellipse.
The center of the ellipse can be found by calculating the midpoint between
the two foci. The midpoint formula is given by
(h, k) = (x1+x2
2,y1+y2
2).
Using F1(−4,0) and F2(4,0), we find
(h, k) = (−4+4
2,0+0
2)= (0,0).
10
Therefore, the center of the ellipse is at the origin, (0,0).
Step 2: Determine the distance between the foci.
The distance between the foci is the major axis of the ellipse, 2a. The
distance formula is given by
2a=√(x2−x1)2+ (y2−y1)2.
Using F1(−4,0) and F2(4,0), we find
2a=√(4 −(−4))2+ (0 −0)2=√64 = 8.
Thus, the major axis of the ellipse is 8 units.
Step 3: Determine the minor axis of the ellipse.
Since the ellipse passes through the point (2,3), the distance from the center
(0,0) to this point is the minor axis, 2b. The distance formula is given by
2b=√(x−h)2+ (y−k)2.
Using (2,3) and (0,0), we find
2b=√(2 −0)2+ (3 −0)2=√13.
Therefore, the minor axis of the ellipse is √13 units.
Step 4: Write the equation of the ellipse.
The equation of an ellipse centered at the origin with major axis 2a along
the x-axis and minor axis 2b along the y-axis is
x2
a2+y2
b2= 1.
Substitute a= 4 and b=√13 into the equation to obtain
x2
16 +y2
13 = 1.
Thus, the equation of the ellipse is x2
16 +y2
13 = 1 .
Question 12
Question
Find the standard form equation of the hyperbola with vertices at (−5,0) and
(5,0) and foci at (−6,0) and (6,0).
11
Solution
Step 1: Find the center of the hyperbola by finding the midpoint between the
vertices:
Center =(−5+5
2,0+0
2)= (0,0)
Step 2: Determine the distance between the center and either vertex to find
a:
a= 5
Step 3: Determine the distance between the center and either focus to find
c:
c= 6
Step 4: Use the relationship between a,band cfor a hyperbola, where bis
the distance between the center and the transverse axis:
c2=a2+b2=⇒b=√c2−a2
b=√62−52=√36 −25 = √11
Step 5: The standard form equation of a hyperbola centered at (h, k)is:
(x−h)2
a2−(y−k)2
b2= 1
Plugging in the values, the standard form equation of the hyperbola is:
x2
25 −y2
11 = 1
Question 13
Question
Find the equation of the hyperbola with vertices at (−3,0) and (3,0), and foci
at (−5,0) and (5,0).
Solution
Step 1: Find the center of the hyperbola.
The center of the hyperbola is the midpoint of the segment connecting the
vertices, which is ((−3 + 3)/2,(0 + 0)/2), or (0,0).
Step 2: Find the distance from the center to either vertex.
The distance from the center to a vertex is the distance from (0,0) to (−3,0)
or (3,0), which is 3.
Step 3: Find the distance between the vertices, which is 2a.
Since the distance between the center and a vertex is 3, we have 2a= 3, so
a= 1.5.
12
Step 4: Find the distance between the foci, which is 2c.
Since the distance between the foci is 10, we have 2c= 10, so c= 5.
Step 5: Use the relationship a2+b2=c2to find b2.
Since a= 1.5and c= 5, we have b2=c2−a2= 52−1.52= 25 −2.25 = 22.75.
Step 6: Write the equation of the hyperbola in standard form.
The equation of a hyperbola with center at (h, k), vertices along the transverse
axis, and foci along the transverse axis is
(x−h)2/a2−(y−k)2/b2= 1
Plugging in the values, the equation becomes
(x−0)2/1.52−(y−0)2/22.75 = 1
Simplifying, we get
x2/2.25 −y2/22.75 = 1
Question 14
Question
Suppose a parabola has a vertex at (−2,3) and its focus is at (−2,5). Write an
equation in standard form for the parabola that opens downward.
Solution
Step 1: Recall the standard form of a parabola that opens downward with vertex
(h, k):
(x−h)2= 4p(y−k)
where the focus is F(h, k +p).
Step 2: We can identify h=−2and k= 3 from the given vertex. Since the
focus is at the point (−2,5), we know that k+p= 5.
Step 3: Solve for pusing the relationship k+p= 5:
3 + p= 5
p= 5 −3
p= 2
Step 4: Substitute h=−2,k= 3, and p= 2 into the standard form of the
equation:
(x+ 2)2= 4 ·2(y−3)
Step 5: Simplify the equation:
(x+ 2)2= 8(y−3)
13
Step 6: Expand the left side of the equation:
x2+ 4x+ 4 = 8y−24
Step 7: Rewrite the equation in standard form by isolating y:
8y=x2+ 4x+ 28
y=1
8x2+1
2x+ 3.5
Step 8: Therefore, the equation of the parabola that opens downward with
the given vertex and focus is:
y=1
8x2+1
2x+ 3.5
Question 15
Question
Solve the following system of equations:
{3x2−y2= 16
x+y= 5
Solution
Step 1: Solve the second equation for one variable in terms of the other. Let’s
solve for xin terms of y:
x= 5 −y
Step 2: Substitute x= 5 −yinto the first equation:
3(5 −y)2−y2= 16
Step 3: Expand and simplify the equation:
75 −30y+ 3y2−y2= 16
2y2−30y+ 59 = 0
Step 4: Solve the quadratic equation for yusing the quadratic formula:
y=−(−30) ±√(−30)2−4(2)(59)
2(2)
y=30 ±√900 −472
4
14
y=30 ±√428
4
y=30 ±2√107
4
y=15 ±√107
2
Step 5: There are two possible values for y:y= 15 + √107/2and y=
15 −√107/2.
Step 6: Substitute each value of yback into the equation x= 5 −yto find
the corresponding values of x: For y= 15 + √107/2:
x= 5 −15 + √107
2=−5 + √107
2
For y= 15 −√107/2:
x= 5 −15 −√107
2=√107 −5
2
Therefore, the solutions to the system of equations are x=−5+√107
2,y=
15 + √107/2and x=√107−5
2,y= 15 −√107/2.
Question 16
Question
Solve the system of equations:
{x2+y2= 4
x2−y2= 1
Solution
Step 1: Start by adding the two equations together to eliminate y2.
x2+y2+x2−y2= 4 + 1
Step 2: Simplify the equation.
2x2= 5
Step 3: Solve for x.
x2=5
2=⇒x=±√5
2
15
Step 4: Substitute the values of xback into the first equation to solve for y.
(√5
2)2+y2= 4
−(√5
2)2+y2= 4
Step 5: Solve for yin both equations.
{5
2+y2= 4
−5
2+y2= 4
Step 6: Simplify the equations.
{y2=3
2
y2=13
2
Step 7: Solve for yin both equations.
y=±√3
2
y=±√13
2
Therefore, the solutions to the system of equations are:
(√5
2,√3
2),(√5
2,−√3
2),(−√5
2,√13
2),(−√5
2,−√13
2)
Question 17
Question
Solve the equation of the parabola given below for y:
4x2−16x+ 3y−11 = 0
Solution
Step 1: First, we need to rearrange the equation in the standard form of the
equation of a parabola. The standard form of a vertical parabola is given by:
(x−h)2= 4p(y−k)
where the vertex of the parabola is at (h, k)and the distance from the vertex
to the focus is p.
Step 2: To rewrite the given equation in the standard form, complete the
square for the xterms:
4x2−16x+ 3y−11 = 0
16
4(x2−4x)+3y−11 = 0
4(x2−4x+4)+3y−11 = 4
4(x−2)2+ 3y−11 = 4
Step 3: Rearrange the equation to solve for y:
4(x−2)2+ 3y−11 = 4
3y=−4(x−2)2+ 15
y=−4
3(x−2)2+ 5
Step 4: Therefore, the equation of the parabola solved for yis:
y=−4
3(x−2)2+ 5
Question 18
Question
Given the equation of a hyperbola in standard form: (x−2)2
9−(y+1)2
16 = 1, find
the coordinates of the center, the vertices, the foci, and the equations of the
asymptotes.
Solution
Step 1: Identify the center of the hyperbola.
The center of the hyperbola is given by (h, k)in the standard form (x−h)2
a2−
(y−k)2
b2= 1. Therefore, in this case, the center is at (2,−1).
Step 2: Find the vertices.
The distance from the center to the vertices in the horizontal direction is a= 3,
while in the vertical direction is b= 4. Therefore, the vertices are at (h±a, k),
which gives the vertices as (2 ±3,−1), or (5,−1) and (−1,−1).
Step 3: Determine the foci.
The distance from the center to the foci is given by c=√a2+b2. Using the
values a= 3 and b= 4, we get c=√9 + 16 = √25 = 5. Hence, the foci are at
(h±c, k), which gives the coordinates as (2 ±5,−1), or (−3,−1) and (7,−1).
Step 4: Find the equations of the asymptotes.
The equations of the asymptotes for a hyperbola centered at (h, k)are given
by y−k=±b
a(x−h). Substituting the given values, we find the equations
as y+ 1 = ±4
3(x−2) and simplifying gives the equations of the asymptotes as
y=±4
3x−11
3.
Therefore, the coordinates of the center are (2,−1), the vertices are (5,−1)
and (−1,−1), the foci are (7,−1) and (−3,−1), and the equations of the asymp-
totes are y=4
3x−11
3and y=−4
3x−11
3.
17
Question 19
Question
Solve the system of equations:
{x2−y2= 9
x+y= 3
Solution
Step 1: Add the second equation to the first equation to eliminate y.
x2−y2+x+y= 9 + 3
x2+x−y2+y= 12
x(x+ 1) −y(y+ 1) = 12
Step 2: Substitute x= 3 −yinto the simplified equation.
(3 −y)(3 −y+ 1) −y(y+ 1) = 12
(3 −y)(4 −y)−y2−y= 12
12 −7y+y2= 12
y2−7y= 0
y(y−7) = 0
Step 3: Solve for yby setting each factor to zero.
y= 0 or y= 7
Step 4: Substitute y= 0 and y= 7 back into the equation x+y= 3 to find
x. For y= 0:
x+ 0 = 3
x= 3
For y= 7:
x+ 7 = 3
x=−4
Therefore, the solutions to the system of equations are (3,0) and (−4,7).
Question 20
Question
Find the standard form equation of a hyperbola with vertices at (−3,2) and
(5,2) and foci at (−7,2) and (9,2).
18
Solution
Step 1: First, let’s determine the center of the hyperbola, which lies at the
midpoint of the segment connecting the vertices. In this case, the center is
C(−3+5
2,2+2
2).
Step 2: The center of the hyperbola is C(1,2).
Step 3: Next, let’s find the distance from the center to one of the vertices.
This distance is the distance ain the standard form equation of a hyperbola.
Step 4: The distance from the center to a vertex is a=|1−(−3)|= 4.
Step 5: The distance from the center to a focus point is the distance cin the
standard form equation of a hyperbola.
Step 6: The distance from the center to a focus is c=| − 7−1|= 8.
Step 7: Recall the relationship between a,b, and cfor a hyperbola: a2+b2=
c2.
Step 8: Substitute aand cinto the equation: 42+b2= 82.
Step 9: Simplify the equation to solve for b:16 + b2= 64.
Step 10: Subtract 16 from both sides to get b2= 48.
Step 11: The standard form equation of a hyperbola is (x−h)2
a2−(y−k)2
b2= 1,
with the center at (h, k). In this case, h= 1,k= 2,a= 4, and b=√48.
Step 12: The standard form equation of the hyperbola is (x−1)2
16 −(y−2)2
48 = 1.
Question 21
Question
Solve the following system of equations:
{x2+y2= 10
x−y= 3
Solution
Let’s solve the system of equations step by step:
Step 1: Solve the second equation for xin terms of y:
x=y+ 3
Step 2: Substitute xin terms of yinto the first equation:
(y+ 3)2+y2= 10
Step 3: Expand and simplify:
y2+ 6y+9+y2= 10
2y2+ 6y−1 = 0
19
Step 4: Solve the quadratic equation for yusing the quadratic formula:
y=−6±√62−4·2·(−1)
2·2
y=−6±√36 + 8
4
y=−6±√44
4
y=−6±2√11
4
y=−3±√11
2
Step 5: Find the corresponding values of xusing x=y+3: For y=−3+√11
2,
we have x=−3+√11
2+ 3
For y=−3−√11
2, we have x=−3−√11
2+ 3
Step 6: So, the solutions to the system of equations are:
(x, y) = (−3 + √11
2+ 3,−3 + √11
2)
(x, y) = (−3−√11
2+ 3,−3−√11
2)
Question 22
Question
Find the standard form of the equation of the ellipse given the following infor-
mation: The ellipse has a major axis of length 10 units and passes through the
points (−3,2) and (7,2).
Solution
Step 1: Find the center of the ellipse. The center of the ellipse is the midpoint
of the major axis. The midpoit of the segment joining (−3,2) and (7,2) is
(−3+7
2,2+2
2) = (2,2).
Therefore, the center of the ellipse is (2,2).
Step 2: Find the endpoints of the major axis. The major axis is along the
x-axis, so the endpoints are (2 −5,2) and (2 + 5,2). This gives the endpoints
(−3,2) and (7,2), as expected.
20
Step 3: Find the equation of the ellipse. Since the major axis is along the
x-axis, the equation of the ellipse is of the form
(x−h)2
a2+(y−k)2
b2= 1,
where (h, k)is the center of the ellipse, 2ais the length of the major axis, and
2bis the length of the minor axis.
Given that the major axis has length 10 units, 2a= 10 so a= 5.
Plugging in the center (2,2) and a= 5 into the equation, we get
(x−2)2
25 +(y−2)2
b2= 1.
Step 4: Find b. Since the ellipse passes through the point (−3,2), we have
(−3−2)2
25 +(2 −2)2
b2= 1,
which simplifies to 25
25 +0
b2= 1,
and leads to 1 = 1 + 0, which is true for any b= 0.
Therefore, the equation of the ellipse is
(x−2)2
25 +(y−2)2
b2= 1 .
Question 23
Question
Let f(x) = −2x2−4x+ 3. Determine the vertex, focus, directrix, and axis of
symmetry of the parabola defined by f(x).
Solution
Step 1: To find the vertex of the parabola defined by f(x), we use the formula
h=−b
2a.
Step 1: h=−b
2a=−−4
2(−2) =−4
−4= 1.
Step 2: Substitute h= 1 into the function f(x)to find the y-coordinate of
the vertex.
Step 2: f(1) = −2(1)2−4(1) + 3 = −2−4 + 3 = −3.
So, the vertex is at (1,−3).
21
Step 3: The axis of symmetry of the parabola is the vertical line passing
through the vertex. In this case, the axis of symmetry is x= 1.
Step 4: To find the focus and directrix of the parabola, we use the formulas
p=1
4aand y=k−p, where kis the y-coordinate of the vertex.
Step 4: p=1
4(−2) =−1
8.
Substitute p=−1
8and k=−3into y=k−pto find the y-coordinate of the
focus and the equation of the directrix.
Step 4: y=−3−(−1
8)=−23
8.
Therefore, the focus is at (1,−23
8)and the directrix has the equation y=−15
8.
Question 24
Question
Find the standard form of the equation of the hyperbola with foci at (−3,0)
and (3,0) and vertices at (−4,0) and (4,0).
Solution
Step 1: Find the center of the hyperbola. The center is the midpoint of the foci,
which is ((−3 + 3)/2,0) = (0,0).
Step 2: Find a, the distance from the center to a vertex. a= 4.
Step 3: Find c, the distance from the center to a focus. c= 3.
Step 4: Use the relationship for hyperbolas: c2=a2+b2to solve for b. Plug
in the values of a= 4 and c= 3.
32= 42+b2
9 = 16 + b2
b2=−7
Since b2<0, this hyperbola is horizontal.
Step 5: The standard form of the equation of a hyperbola with a horizontal
transverse axis and center at the origin is:
x2
a2−y2
b2= 1
Therefore, the standard form of the equation of the hyperbola is:
x2
16 −y2
7= 1
22
Question 25
Question
Find the standard form of the equation of the parabola that satisfies the given
conditions: vertex at (−1,3) and focus at (−1,1).
Solution
Step 1: Recall that the standard form of the equation of a parabola with vertex
at (h, k)and focus at (h, k +p)or (h+p, k)is given by:
(x−h)2= 4p(y−k)
where pis the distance from the vertex to the focus.
Step 2: In this case, the vertex is (−1,3) and the focus is (−1,1). This
means that p= 3 −1 = 2.
Step 3: Substituting the values of h,k, and pinto the standard form equa-
tion, we have:
(x+ 1)2= 8(y−3)
Therefore, the standard form of the equation of the parabola is (x+ 1)2=
8(y−3).
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