MATH 121 - COLLEGE ALGEBRA -
Equations and properties of conic
sections
Question Bank - Set 2
Liberty University
Question 1
Question
Determine the equation of a hyperbola with vertices at (−3,0) and (3,0) and
passing through the point (4,2).
Solution
Step 1: Find the center of the hyperbola. The center is the midpoint of the
segment connecting the vertices. The coordinates of the center are given by the
formula:
(h, k) = ((−3 + 3)
2,(0 + 0)
2)= (0,0).
Step 2: Find the distance between the center and one of the vertices, which
is the distance from the center to either of the foci. Since the vertices are 3
units apart on the x-axis, the distance between the center and either vertex (or
focus) is 3 units. Now we have the value of a= 3.
Step 3: Determine the distance between the center and the point (4,2). This
will give us the value of c. The distance is given by the equation:
c=√(x2−x1)2+ (y2−y1)2=√(4 −0)2+ (2 −0)2=√16 + 4 = √20.
Step 4: Use the relationship a2+b2=c2to find b, where bis the distance
from the center to the asymptotes.
32+b2=√202
9 + b2= 20
b2= 11
b=√11.
Step 5: Write the equation of the hyperbola in standard form (x2
a2−y2
b2= 1).
The equation of the hyperbola is therefore:
x2
9−y2
11 = 1.
Question 2
Question
Find the standard form equation of a parabola with a focus at (−3,2) and a
directrix given by y=−4.
Solution
Step 1: Recall that the standard form equation of a parabola with a horizontal
axis of symmetry and a focus at (h+k, k)is given by
(x−h)2= 4p(y−k),
where pis the distance from the vertex to the focus.
Step 2: In this problem, we know that the focus is at (−3,2), so h=−3and
k= 2.
Step 3: The given directrix is y=−4, which means the distance from the
vertex to the directrix is |k−(−4)|=|2+4|= 6.
Step 4: Since the parabola opens upwards from the directrix, p= 6.
Step 5: Now, we can plug in the values of h,k, and pinto the standard form
equation to find the equation of the parabola.
(x+ 3)2= 4 ·6(y−2)
Step 6: Simplify the equation to get the standard form of the parabola.
(x+ 3)2= 24(y−2)
Step 7: Therefore, the standard form equation of the parabola with a focus
at (−3,2) and a directrix given by y=−4is (x+ 3)2= 24(y−2).
Question 3
Question
Given the equation of a hyperbola as 4x2−9y2= 36, find the center, vertices,
foci, and equations of the asymptotes.
2
Solution
Step 1: Rewrite the given equation in standard form by dividing each term on
both sides by 36.
x2
9−y2
4= 1
Step 2: Identify the center of the hyperbola, which is at the point (h, k) =
(0,0).
Step 3: Determine the vertices of the hyperbola. Since the hyperbola opens
horizontally, the vertices are located at (h±a, k), where a= 3. So, the vertices
are at (−3,0) and (3,0).
Step 4: Find the foci of the hyperbola. The distance from the center to each
focus is given by c=√a2+b2. Substitute a= 3 and b= 2 into the formula to
find c.
c=√32+ 22=√13
Therefore, the foci are located at (√13,0) and (−√13,0).
Step 5: Calculate the equations of the asymptotes. The general form for
the asymptotes is y=±b
ax. Substitute a= 3 and b= 2 into the equation to
find the equations of the asymptotes. So, the equations of the asymptotes are
y=±2
3x.
Question 4
Question
Solve the following system of equations:
{2x2−3y2= 1
3x2−2y2=−1
Solution
Step 1: Rewrite the system of equations in standard form by eliminating y2.
{2x2−3y2= 1
3x2−2y2=−1
To eliminate y2, we will multiply the first equation by 2 and the second
equation by 3:
{4x2−6y2= 2
9x2−6y2=−3
Step 2: Subtract the second equation from the first equation to eliminate
y2.
(4x2−6y2)−(9x2−6y2) = 2 −(−3)
3
Simplify to get:
−5x2= 5
Step 3: Solve for x. Divide both sides by −5to solve for x:
x2=−1
Since x2=−1has no real solutions, the system of equations has no real solu-
tions.
Question 5
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: Determine the center of the hyperbola. Since the center of a hyperbola
is the midpoint of the line segment connecting the vertices, the center of the
hyperbola is (0,0).
Step 2: Find the distance between the center and a vertex to determine a.
The distance between the center (0,0) and either of the vertices (3,0) or
(−3,0) is 3 units.
So a= 3.
Step 3: Find the distance between the center and a focus to determine c.
The distance between the center (0,0) and either of the foci (5,0) or (−5,0)
is 5 units.
So c= 5.
Step 4: Use the relationship c2=a2+b2to find b.
Since we have a= 3 and c= 5, we can substitute these values into the
equation to solve for b:
52= 32+b2
25 = 9 + b2
b2= 16
b=±4
Step 5: Write the equation of the hyperbola in standard form.
The equation of a hyperbola centered at the origin with vertices on the x-axis
is x2
a2−y2
b2= 1.
Since a= 3,b= 4, the equation of the hyperbola is:
x2
9−y2
16 = 1
4
Question 6
Question
Find the standard form equation of a hyperbola with vertices at (−2,1) and
(4,1), and the distance between its foci is √29.
Solution
Step 1: First, we find the center of the hyperbola by finding the midpoint of the
vertices. The midpoint formula is given by
(x1+x2
2,y1+y2
2).
Step 2: Applying the midpoint formula, we find the center of the hyperbola:
(−2+4
2,1+1
2)= (1,1).
Step 3: Since the distance between the vertices is 6units, we can determine
a, the distance from the center to a vertex, which is 3units.
Step 4: Given that the distance between the foci is √29, and that for a
hyperbola c2=a2+b2, where cis the distance between the foci, we have
c2= 29. Thus, c=√29.
Step 5: We know that c2=a2+b2, where a= 3. Plugging in the values, we
get
29 = 9 + b2,
b2= 29 −9 = 20.
Step 6: Now, we can write the standard form equation of the hyperbola
centered at (1,1):
(x−1)2
9−(y−1)2
20 = 1.
Therefore, the standard form equation of the hyperbola is (x−1)2
9−(y−1)2
20 =
1.
Question 7
Question
Solve the equation 3x2−12x−9y2+ 36y= 0 for yin terms of x.
5
Solution
Step 1: Rearrange the equation to isolate the terms involving y:
3x2−12x= 9y2−36y
Step 2: Factor out the coefficients in front of y2and y:
3(x2−4x) = 9(y2−4y)
Step 3: Complete the square on the right side of the equation by adding
(4/2)2= 4 to both sides:
3(x2−4x+ 4) = 9(y2−4y+ 4)
Step 4: Simplify both sides of the equation:
3(x−2)2= 9(y−2)2
Step 5: Divide both sides by 9 to solve for y:
(x−2)2= 3(y−2)2
Step 6: Take the square root of both sides:
√(x−2)2=±√3(y−2)2
Step 7: Simplify:
|x−2|=√3|y−2|
Step 8: Solve for yby isolating it on one side of the equation:
|y−2|=1
√3|x−2|
Therefore, the equation in terms of yis |y−2|=1
√3|x−2|.
Question 8
Question
Solve the system of equations:
{x2+y2= 25
xy = 12
6
Solution
Step 1: Let’s solve the second equation for yin terms of x.
xy = 12 =⇒y=12
x
Step 2: Substitute y=12
xinto the first equation.
x2+(12
x)2
= 25
Step 3: Simplify the equation.
x2+144
x2= 25
Step 4: Multiply through by x2to clear the fraction.
x4+ 144 = 25x2
Step 5: Rearrange the equation to form a quadratic equation.
x4−25x2+ 144 = 0
Step 6: Letting u=x2, the equation becomes
u2−25u+ 144 = 0
Step 7: Solve the quadratic equation by factoring or using the quadratic
formula. The factors of 144 that add up to -25 are -16 and -9. So, the equation
factors as (u−16)(u−9) = 0.
Step 8: Solve for u.
u= 16 or u= 9
Step 9: Substitute u=x2back in.
x2= 16 or x2= 9
Step 10: Solve for x.
x=±4or x=±3
Step 11: Substitute xinto y=12
xto find the corresponding values of y. For
x= 4,y=12
4= 3 or y=−3. For x=−4,y=−3or y= 3. For x= 3,
y=12
3= 4 or y=−4. For x=−3,y=−4or y= 4.
Step 12: Therefore, the solutions to the system are (−4,−3),(−4,3),(4,−3),
and (4,3).
7
Question 9
Question
Solve the following system of equations:
{x2+y2= 25
x−y= 3
Solution
Step 1: Rewrite the second equation as 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 for yusing the quadratic formula:
y=−6±√62−4(2)(−16)
2(2)
y=−6±√36 + 128
4
y=−6±√164
4
y=−6±2√41
4
y=−3±√41
2
Step 5: Substitute the values of yback into the equation x=y+ 3 to find
the corresponding values of x:
x=−3 + √41
2+ 3 or x=−3−√41
2+ 3
x=−3 + √41 + 6
2or x=−3−√41 + 6
2
x=3 + √41
2or x=3−√41
2
Step 6: Therefore, the solutions to the system of equations are:
(x, y) = (3 + √41
2,−3 + √41
2)and (3−√41
2,−3−√41
2)
8
Question 10
Question
Solve the equation 4x2+ 9y2−24x+ 54y−45 = 0 and determine the type of
conic section represented by this equation.
Solution
Step 1: Rearrange the equation by completing the square for both xand y
terms.
4x2+ 9y2−24x+ 54y−45 = 0
4(x2−6x) + 9(y2+ 6y) = 45
4(x2−6x+ 9) + 9(y2+ 6y+ 9) = 45 + 36 + 81
4(x−3)2+ 9(y+ 3)2= 162
Step 2: Divide by 162 to normalize the equation.
4(x−3)2
162 +9(y+ 3)2
162 = 1
(x−3)2
162
4
+(y+ 3)2
162
9
= 1
(x−3)2
40.5+(y+ 3)2
18 = 1
Step 3: Compare the equation with the standard form of conic sections to
determine the type of conic section. The given equation matches the standard
form of an ellipse:
((x−h)2
a2)+((y−k)2
b2) = 1
where the center C(h, k), major axis length 2a, and minor axis length 2b. In
this case, C(3,−3),a=√40.5, and b=√18, so the conic section represented
by the equation is an ellipse.
Question 11
Question
Solve the following system of equations and identify the type of conic section
represented by the graph of the equations: {x2+y2−4x−6y+ 9 = 0
3x2−2y2−12x+ 8y−19 = 0
9
Solution
Step 1: Complete the square for the first equation by rearranging terms:
x2−4x+y2−6y=−9
Adding and subtracting the necessary constants to complete the square on
both xand yterms:
x2−4x+4+y2−6y+ 9 = −9+4+9
(x−2)2+ (y−3)2= 4
Step 2: Complete the square for the second equation by rearranging terms:
3x2−12x−2y2+ 8y= 19
Adding and subtracting the necessary constants to complete the square on
both xand yterms:
3(x2−4x)−2(y2−4y) = 19
3(x2−4x+ 4) −2(y2−4y+ 4) = 19 + 3(4) −2(4)
3(x−2)2−2(y−2)2= 27
(x−2)2
(3
√27 )2−(y−2)2
(2
√27 )2= 1
Step 3: Comparing the equations with the standard forms of conic sections:
The first equation represents a circle with center at (2, 3) and radius 2. The
second equation represents a hyperbola with center at (2, 2), horizontal axis
length 2√27, and vertical axis length 2√27/√2.
Therefore, the system of equations represents a circle and a hyperbola.
Question 12
Question
Let C(x) = 3x2−4x−10 be the cost function and R(x) = 4x+ 15 be the
revenue function for a company. Find the break-even point(s) for this company,
if they exist.
Solution
Step 1: To find the break-even point, we need to set the cost equal to the revenue
and solve for x. This means we need to solve the equation C(x) = R(x).
Step 2: First, we substitute the given functions into the equation:
3x2−4x−10 = 4x+ 15
10
Step 3: Next, we simplify the equation by moving all terms to one side:
3x2−4x−10 −4x−15 = 0
Step 4: Combine like terms:
3x2−8x−25 = 0
Step 5: Now, we need to solve the quadratic equation 3x2−8x−25 = 0.
We can do this by factoring or using the quadratic formula.
Step 6: Factoring the quadratic equation gives us:
(3x+ 5)(x−5) = 0
Step 7: Setting each factor equal to zero gives us the solutions:
3x+ 5 = 0 or x−5 = 0
Step 8: Solving these equations gives us:
x=−5
3or x= 5
Step 9: Therefore, the break-even points for the company are x=−5
3and
x= 5.
Question 13
Question
Solve the equation 5x2+ 12xy + 8y2−14x−34y−19 = 0 for yin terms of x.
Solution
Step 1: Rewrite the given equation in a standard form of a conic section.
Step 2: Complete the square for the quadratic terms to find the standard
form.
Step 3: Factor the quadratic expression and solve for yin terms of x.
Step 1: Rewrite the equation in a standard form of a conic section. We have
the equation:
5x2+ 12xy + 8y2−14x−34y−19 = 0
Step 2: Complete the square for the quadratic terms. To complete the square
for the xand yterms, we need to group the xterms, yterms, and constant
terms separately. The quadratic terms can be rewritten as:
(5x2+ 12xy + 8y2)−14x−34y−19 = 0
11
Now, complete the square for the quadratic terms (5x2+ 12xy + 8y2). We
have:
5(x2+12
5xy)+8y2−14x−34y−19 = 0
Step 3: Factor the quadratic expression. We can rewrite the quadratic part
as a perfect square trinomial:
5(x+6
5y)2
−36
5y2−14x−34y−19 = 0
Factor out the negative sign and simplify:
5(x+6
5y)2
−36
5(y2+5
2y)−14x−19 = 0
5(x+6
5y)2
−36
5(y2+5
2y+25
4)−14x−19 = 0
5(x+6
5y)2
−36
5(y+5
2)2−14x−19 = 0
Now, we can rewrite the equation in the standard form:
5(x+6
5y)2
−36
5(y+5
2)2= 14x+ 19
Therefore, the equation in terms of yis
36
5(y+5
2)2= 5(x+6
5y)2−14x−19
Now, solve for yin terms of x:
y+5
2=±√5
36(x+6
5y)2−14x+ 19
36
y=−5
2±√5
36(x+6
5y)2−14x+ 19
36
Question 14
Question
Find the standard form equation of the ellipse that satisfies the given conditions:
The ellipse has a major axis along the y-axis, a minor axis of length 6, and passes
through the points (0,5) and (0,−5).
12
Solution
Step 1: Since the major axis is along the y-axis, the equation of the ellipse will
be of the form y2
a2+x2
b2= 1, where ais the length of the semi-major axis and b
is the length of the semi-minor axis.
Step 2: Since the ”major axis” is the vertical axis, the length of the major
axis is 2a, and therefore 2a= 12 (from the given minor axis length of 6).
Step 3: Solving for a, we get a= 6.
Step 4: The center of the ellipse lies on the y-axis, so the x-coordinate of the
center is 0. Let’s denote the y-coordinate of the center as k.
Step 5: Using the midpoint formula, we find the center of the ellipse:
k=5+(−5)
2= 0
So, the center of the ellipse is at (0,0).
Step 6: We can now write the equation of the ellipse as:
y2
36 +x2
9= 1
This is the standard form equation of the ellipse.
Question 15
Question
Solve the following system of equations:
{x2+y2= 25
2x−y= 1
Solution
Step 1: Solve the second equation for yin terms of x:
2x−y= 1 =⇒y= 2x−1
Step 2: Substitute y= 2x−1into the first equation:
x2+ (2x−1)2= 25
Step 3: Expand and simplify the equation:
x2+ 4x2−4x+ 1 = 25
5x2−4x−24 = 0
13
Step 4: Factor the quadratic equation:
(5x+ 6)(x−4) = 0
Step 5: Set each factor equal to zero and solve for x:
5x+ 6 = 0 =⇒x=−6
5
x−4 = 0 =⇒x= 4
Step 6: Substitute x=−6
5and x= 4 back into y= 2x−1to find the
corresponding values of y: For x=−6
5:
y= 2 (−6
5)−1 = −12
5−1 = −17
5
For x= 4:
y= 2(4) −1 = 8 −1 = 7
Step 7: Therefore, the solution to the system of equations is:
{x=−6
5, y =−17
5
x= 4, y = 7
Question 16
Question
Find the standard form equation of the ellipse that satisfies the following con-
ditions: Center at (−3,2), major axis along the y-axis with a length of 8, and
minor axis along the x-axis with a length of 6.
Solution
Step 1: The general form of the equation of an ellipse centered at (h, k)with
major axis of length 2aalong the x-axis and minor axis of length 2balong the
y-axis is given by:
(x−h)2
a2+(y−k)2
b2= 1
Step 2: Since the center of the ellipse is (−3,2), we have h=−3and k= 2.
Step 3: The major axis length is 8, so 2a= 8 which gives a= 4.
Step 4: The minor axis length is 6, so 2b= 6 which gives b= 3.
Step 5: Now, substitute the values of h,k,a, and binto the equation of the
ellipse:
(x+ 3)2
42+(y−2)2
32= 1
14
Step 6: Simplify the equation:
(x+ 3)2
16 +(y−2)2
9= 1
Therefore, the standard form equation of the ellipse is (x+ 3)2
16 +(y−2)2
9= 1 .
Question 17
Question
Solve the system of equations:
{x2−y2= 9
2x+ 2y= 0
Solution
Step 1: Rearrange the second equation to solve for yin terms of x.
2x+ 2y= 0 ⇒2y=−2x⇒y=−x
Step 2: Substitute y=−xinto the first equation and solve for x.
x2−(−x)2= 9 ⇒x2−x2= 9 ⇒0 = 9
Step 3: Since 0= 9, the system of equations has no solution.
Question 18
Question
Solve the following system of equations:
{2x2+ 5y2−4x−20y−1 = 0
3x2−y2+ 2x−4y+ 2 = 0
Solution
Step 1: We will begin by completing the square for both equations. For the first
equation, we rewrite it as:
2(x2−2x) + 5(y2−4y) = 1
We complete the square for both xand yterms separately:
2(x2−2x+ 1) + 5(y2−4y+ 4) = 1 + 2 + 20
15
Simplifying gives:
2(x−1)2+ 5(y−2)2= 23
Step 2: Following the same process, we rewrite the second equation as:
3(x2+2
3x)−(y2−4y) = −2
Completing the square for both xand yterms separately:
3(x2+2
3x+(1
3)2)−(y2−4y+ 4) = −2+1+4
Simplifying gives:
3(x+1
3)2
−(y−2)2= 3
Step 3: Now we have the system in standard form:
{2(x−1)2+ 5(y−2)2= 23
3(x+1
3)2−(y−2)2= 3
Step 4: This system represents the equations of an ellipse and a hyperbola.
The solution to the system is the intersection points of these two conics. By
substituting the second equation into the first, or vice versa, we can solve for x
and y.
Unfortunately, the algebra required to solve this system of equations exceeds
the complexity of a typical College Algebra course.
Question 19
Question
Solve the system of equations:
{3x2−y2= 4
x+y= 2
Solution
Step 1: Let’s first solve the second equation for xin terms of y.
x+y= 2
x= 2 −y
16
Step 2: Substitute x= 2 −yinto the first equation.
3(2 −y)2−y2= 4
3(4 −4y+y2)−y2= 4
12 −12y+ 3y2−y2= 4
12 −12y+ 2y2= 4
2y2−12y+ 8 = 0
y2−6y+ 4 = 0
Step 3: Solve the quadratic equation y2−6y+ 4 = 0 using the quadratic
formula.
y=−(−6) ±√(−6)2−4(1)(4)
2(1)
y=6±√36 −16
2
y=6±√20
2
y= 3 ±2√5
Step 4: Find the corresponding values of xfor each solution of y.
For y= 3 + 2√5 :
x= 2 −(3 + 2√5) = −1−2√5
For y= 3 −2√5 :
x= 2 −(3 −2√5) = −1+2√5
Step 5: Therefore, the solutions to the system of equations are:
{x=−1−2√5, y = 3 + 2√5
x=−1+2√5, y = 3 −2√5
Question 20
Question
Solve the system of equations:
{x2−y2= 4
xy = 1
Solution
Step 1: Let’s solve for yin the second equation xy = 1:
y=1
x
17
Step 2: Substitute y=1
xinto the first equation x2−y2= 4:
x2−(1
x)2
= 4
Step 3: Simplify the equation:
x2−1
x2= 4
Step 4: Multiply through by x2to clear the fraction:
x4−1 = 4x2
Step 5: Rearrange the equation:
x4−4x2−1 = 0
Step 6: This is a quadratic equation in terms of x2. Let u=x2:
u2−4u−1 = 0
Step 7: Solve this quadratic equation for uusing the quadratic formula:
u=4±√42−4(1)(−1)
2(1)
u=4±√16 + 4
2
u=4±√20
2
u=4±2√5
2
u= 2 ±√5
Step 8: Since u=x2, there are two possible solutions for x:x2= 2 + √5
and x2= 2 −√5.
Step 9: Solve for x: For x2= 2 + √5,x=√2 + √5or x=−√2 + √5. For
x2= 2 −√5,x=√2−√5or x=−√2−√5.
Step 10: Substituting the values of xback into y=1
xto find the corre-
sponding values of y: For x=√2 + √5,y=1
√2+√5or y=−1
√2+√5. For
x=−√2 + √5,y=−1
√2+√5or y=1
√2+√5. Similarly for the other two
values of x.
Therefore, the solutions to the system of equations are:
(x, y) = (√2 + √5,1
√2 + √5),(−√2 + √5,−1
√2 + √5),(√2−√5,1
√2−√5),(−√2−√5,−1
√2−√5)
18
Question 21
Question
Given the equation of a parabolic reflector: y=x2−4, find the focus and
directrix.
Solution
Step 1: To find the vertex of the parabolic reflector, rewrite the equation in
vertex form. The vertex form of a parabolic equation is y=a(x−h)2+k,
where (h, k)is the vertex. Step 2: Compare the given equation with the vertex
form and identify hand k. Step 3: The vertex of the parabolic reflector is at
(h, k). Step 4: The focus of a parabolic reflector is given by the point (h, k+1
4a).
Step 5: The directrix of a parabolic reflector is the line y=k−1
4a.
Therefore, the focus and directrix of the parabolic reflector y=x2−4are
found by substituting a= 1,h= 0, and k=−4into the formulas for the focus
and directrix.
Question 22
Question
Let y=−3x2+ 4x−2be the equation of a parabola. Find the coordinates of
the vertex, the focus, and the directrix of this parabola.
Solution
Step 1: To find the vertex of the parabola, use the formula x=−b
2ato find the
x-coordinate of the vertex. Then substitute this x value back into the equation
to find the y-coordinate.
Step 1: Find the x-coordinate of the vertex.
The quadratic equation is in the form y=ax2+bx +c, so a=−3,b= 4.
Substitute these values into x=−b
2a.
x=−4
2(−3) =2
3
Step 2: Find the y-coordinate of the vertex.
Plug x=2
3back into the equation y=−3x2+ 4x−2.
y=−3(2
3)2
+ 4 (2
3)−2
y=−2
19
Thus, the vertex of the parabola is (2
3,−2).
Step 2: To find the focus of the parabola, use the formula p=1
4awhere pis
the distance from the vertex to the focus.
Step 3: Find the distance from the vertex to the focus.
p=1
4(−3) =−1
12 (Note: The negative sign indicates that the parabola opens
downwards)
Step 4: Find the focus.
Since the parabola opens downwards, the focus is at a distance pbelow the
vertex. The y-coordinate of the focus is −2−1
12 =−25
12 .
Therefore, the focus of the parabola is (2
3,−25
12 ).
Step 3: To find the equation of the directrix, use the formula y=k+1
4a
where kis the y-coordinate of the vertex.
Step 5: Find the equation of the directrix.
y=−2 + 1
4(−3) =−2−1
12 =−25
12
Hence, the equation of the directrix is y=−25
12 .
Question 23
Question
Solve the following system of equations:
{x2+y2= 25
x−y= 1
Solution
Step 1: Solve the second equation for xin terms of y:
x=y+ 1
Step 2: Substitute x=y+ 1 into the first equation:
(y+ 1)2+y2= 25
Step 3: Expand the left side of the equation:
y2+ 2y+1+y2= 25
Step 4: Combine like terms:
2y2+ 2y−24 = 0
20
Step 5: Divide the equation by 2 to simplify it,
y2+y−12 = 0
Step 6: Factor the quadratic equation:
(y+ 4)(y−3) = 0
Step 7: Solve for y,
y=−4,3
Step 8: Substitute the values of yback into x=y+1 to find the correspond-
ing values of x:
x=−4 + 1 = −3, x = 3 + 1 = 4
Step 9: The solutions to the system of equations are (−3,−4) and (4,3).
Question 24
Question
Find the standard form of the equation of the hyperbola with foci at (−5,0)
and (5,0) and vertices at (−7,0) and (7,0).
Solution
Step 1: Identify the center of the hyperbola by finding the midpoint of the
segment connecting the vertices. The center is the point (h, k).
Midpoint =(−7+7
2,0+0
2)= (0,0)
Step 2: Determine the value of aby calculating the distance from the center
to one of the vertices. In this case, a= 7.
Step 3: Find the value of c, which is the distance from the center to one of
the foci.
c= 5
Step 4: Use the relationship c2=a2+b2to solve for b.
25 = 49 + b2
b2= 25 −49 = −24
Step 5: Write the equation of the hyperbola in standard form (x−h)2
a2−
(y−k)2
b2= 1.
x2
49 −y2
−24 = 1
Therefore, the standard form of the equation of the hyperbola is x2
49 +y2
24 = 1.
21
Question 25
Question
Find the standard form of the equation of a circle passing through the points
(1,2) and (5,6).
Solution
Step 1: Use the general equation of a circle in standard form: (x−h)2+(y−k)2=
r2, where (h, k)is the center of the circle and ris the radius.
Step 2: Find the midpoint of the given points to get the center of the circle
(h, k). Midpoint formula: (x1+x2
2,y1+y2
2).
Plugging in the points (1,2) and (5,6):h=1+5
2= 3, and k=2+6
2= 4, so
the center is at (3,4).
Step 3: Calculate the radius rby substituting one of the given points into the
equation of the circle and solving for r. We will use (1,2).(1−3)2+(2−4)2=r2
(−2)2+ (−2)2=r24 + 4 = r2r2= 8 r=√8 = 2√2.
Step 4: Substitute the center and radius into the standard form of the equa-
tion of a circle. (x−3)2+ (y−4)2= (2√2)2(x−3)2+ (y−4)2= 8
Therefore, the standard form of the equation of the circle passing through
the points (1,2) and (5,6) is (x−3)2+ (y−4)2= 8.
22
b2= 11
b=√11.
Step 5: Write the equation of the hyperbola in standard form (x2
a2−y2
b2= 1).
The equation of the hyperbola is therefore:
x2
9−y2
11 = 1.
Question 2
Question
Find the standard form equation of a parabola with a focus at (−3,2) and a
directrix given by y=−4.
Solution
Step 1: Recall that the standard form equation of a parabola with a horizontal
axis of symmetry and a focus at (h+k, k)is given by
(x−h)2= 4p(y−k),
where pis the distance from the vertex to the focus.
Step 2: In this problem, we know that the focus is at (−3,2), so h=−3and
k= 2.
Step 3: The given directrix is y=−4, which means the distance from the
vertex to the directrix is |k−(−4)|=|2+4|= 6.
Step 4: Since the parabola opens upwards from the directrix, p= 6.
Step 5: Now, we can plug in the values of h,k, and pinto the standard form
equation to find the equation of the parabola.
(x+ 3)2= 4 ·6(y−2)
Step 6: Simplify the equation to get the standard form of the parabola.
(x+ 3)2= 24(y−2)
Step 7: Therefore, the standard form equation of the parabola with a focus
at (−3,2) and a directrix given by y=−4is (x+ 3)2= 24(y−2).
Question 3
Question
Given the equation of a hyperbola as 4x2−9y2= 36, find the center, vertices,
foci, and equations of the asymptotes.
2
Solution
Step 1: Rewrite the given equation in standard form by dividing each term on
both sides by 36.
x2
9−y2
4= 1
Step 2: Identify the center of the hyperbola, which is at the point (h, k) =
(0,0).
Step 3: Determine the vertices of the hyperbola. Since the hyperbola opens
horizontally, the vertices are located at (h±a, k), where a= 3. So, the vertices
are at (−3,0) and (3,0).
Step 4: Find the foci of the hyperbola. The distance from the center to each
focus is given by c=√a2+b2. Substitute a= 3 and b= 2 into the formula to
find c.
c=√32+ 22=√13
Therefore, the foci are located at (√13,0) and (−√13,0).
Step 5: Calculate the equations of the asymptotes. The general form for
the asymptotes is y=±b
ax. Substitute a= 3 and b= 2 into the equation to
find the equations of the asymptotes. So, the equations of the asymptotes are
y=±2
3x.
Question 4
Question
Solve the following system of equations:
{2x2−3y2= 1
3x2−2y2=−1
Solution
Step 1: Rewrite the system of equations in standard form by eliminating y2.
{2x2−3y2= 1
3x2−2y2=−1
To eliminate y2, we will multiply the first equation by 2 and the second
equation by 3:
{4x2−6y2= 2
9x2−6y2=−3
Step 2: Subtract the second equation from the first equation to eliminate
y2.
(4x2−6y2)−(9x2−6y2) = 2 −(−3)
3
Simplify to get:
−5x2= 5
Step 3: Solve for x. Divide both sides by −5to solve for x:
x2=−1
Since x2=−1has no real solutions, the system of equations has no real solu-
tions.
Question 5
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: Determine the center of the hyperbola. Since the center of a hyperbola
is the midpoint of the line segment connecting the vertices, the center of the
hyperbola is (0,0).
Step 2: Find the distance between the center and a vertex to determine a.
The distance between the center (0,0) and either of the vertices (3,0) or
(−3,0) is 3 units.
So a= 3.
Step 3: Find the distance between the center and a focus to determine c.
The distance between the center (0,0) and either of the foci (5,0) or (−5,0)
is 5 units.
So c= 5.
Step 4: Use the relationship c2=a2+b2to find b.
Since we have a= 3 and c= 5, we can substitute these values into the
equation to solve for b:
52= 32+b2
25 = 9 + b2
b2= 16
b=±4
Step 5: Write the equation of the hyperbola in standard form.
The equation of a hyperbola centered at the origin with vertices on the x-axis
is x2
a2−y2
b2= 1.
Since a= 3,b= 4, the equation of the hyperbola is:
x2
9−y2
16 = 1
4
Question 6
Question
Find the standard form equation of a hyperbola with vertices at (−2,1) and
(4,1), and the distance between its foci is √29.
Solution
Step 1: First, we find the center of the hyperbola by finding the midpoint of the
vertices. The midpoint formula is given by
(x1+x2
2,y1+y2
2).
Step 2: Applying the midpoint formula, we find the center of the hyperbola:
(−2+4
2,1+1
2)= (1,1).
Step 3: Since the distance between the vertices is 6units, we can determine
a, the distance from the center to a vertex, which is 3units.
Step 4: Given that the distance between the foci is √29, and that for a
hyperbola c2=a2+b2, where cis the distance between the foci, we have
c2= 29. Thus, c=√29.
Step 5: We know that c2=a2+b2, where a= 3. Plugging in the values, we
get
29 = 9 + b2,
b2= 29 −9 = 20.
Step 6: Now, we can write the standard form equation of the hyperbola
centered at (1,1):
(x−1)2
9−(y−1)2
20 = 1.
Therefore, the standard form equation of the hyperbola is (x−1)2
9−(y−1)2
20 =
1.
Question 7
Question
Solve the equation 3x2−12x−9y2+ 36y= 0 for yin terms of x.
5
Solution
Step 1: Rearrange the equation to isolate the terms involving y:
3x2−12x= 9y2−36y
Step 2: Factor out the coefficients in front of y2and y:
3(x2−4x) = 9(y2−4y)
Step 3: Complete the square on the right side of the equation by adding
(4/2)2= 4 to both sides:
3(x2−4x+ 4) = 9(y2−4y+ 4)
Step 4: Simplify both sides of the equation:
3(x−2)2= 9(y−2)2
Step 5: Divide both sides by 9 to solve for y:
(x−2)2= 3(y−2)2
Step 6: Take the square root of both sides:
√(x−2)2=±√3(y−2)2
Step 7: Simplify:
|x−2|=√3|y−2|
Step 8: Solve for yby isolating it on one side of the equation:
|y−2|=1
√3|x−2|
Therefore, the equation in terms of yis |y−2|=1
√3|x−2|.
Question 8
Question
Solve the system of equations:
{x2+y2= 25
xy = 12
6
Solution
Step 1: Let’s solve the second equation for yin terms of x.
xy = 12 =⇒y=12
x
Step 2: Substitute y=12
xinto the first equation.
x2+(12
x)2
= 25
Step 3: Simplify the equation.
x2+144
x2= 25
Step 4: Multiply through by x2to clear the fraction.
x4+ 144 = 25x2
Step 5: Rearrange the equation to form a quadratic equation.
x4−25x2+ 144 = 0
Step 6: Letting u=x2, the equation becomes
u2−25u+ 144 = 0
Step 7: Solve the quadratic equation by factoring or using the quadratic
formula. The factors of 144 that add up to -25 are -16 and -9. So, the equation
factors as (u−16)(u−9) = 0.
Step 8: Solve for u.
u= 16 or u= 9
Step 9: Substitute u=x2back in.
x2= 16 or x2= 9
Step 10: Solve for x.
x=±4or x=±3
Step 11: Substitute xinto y=12
xto find the corresponding values of y. For
x= 4,y=12
4= 3 or y=−3. For x=−4,y=−3or y= 3. For x= 3,
y=12
3= 4 or y=−4. For x=−3,y=−4or y= 4.
Step 12: Therefore, the solutions to the system are (−4,−3),(−4,3),(4,−3),
and (4,3).
7
Question 9
Question
Solve the following system of equations:
{x2+y2= 25
x−y= 3
Solution
Step 1: Rewrite the second equation as 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 for yusing the quadratic formula:
y=−6±√62−4(2)(−16)
2(2)
y=−6±√36 + 128
4
y=−6±√164
4
y=−6±2√41
4
y=−3±√41
2
Step 5: Substitute the values of yback into the equation x=y+ 3 to find
the corresponding values of x:
x=−3 + √41
2+ 3 or x=−3−√41
2+ 3
x=−3 + √41 + 6
2or x=−3−√41 + 6
2
x=3 + √41
2or x=3−√41
2
Step 6: Therefore, the solutions to the system of equations are:
(x, y) = (3 + √41
2,−3 + √41
2)and (3−√41
2,−3−√41
2)
8
Question 10
Question
Solve the equation 4x2+ 9y2−24x+ 54y−45 = 0 and determine the type of
conic section represented by this equation.
Solution
Step 1: Rearrange the equation by completing the square for both xand y
terms.
4x2+ 9y2−24x+ 54y−45 = 0
4(x2−6x) + 9(y2+ 6y) = 45
4(x2−6x+ 9) + 9(y2+ 6y+ 9) = 45 + 36 + 81
4(x−3)2+ 9(y+ 3)2= 162
Step 2: Divide by 162 to normalize the equation.
4(x−3)2
162 +9(y+ 3)2
162 = 1
(x−3)2
162
4
+(y+ 3)2
162
9
= 1
(x−3)2
40.5+(y+ 3)2
18 = 1
Step 3: Compare the equation with the standard form of conic sections to
determine the type of conic section. The given equation matches the standard
form of an ellipse:
((x−h)2
a2)+((y−k)2
b2) = 1
where the center C(h, k), major axis length 2a, and minor axis length 2b. In
this case, C(3,−3),a=√40.5, and b=√18, so the conic section represented
by the equation is an ellipse.
Question 11
Question
Solve the following system of equations and identify the type of conic section
represented by the graph of the equations: {x2+y2−4x−6y+ 9 = 0
3x2−2y2−12x+ 8y−19 = 0
9
Solution
Step 1: Complete the square for the first equation by rearranging terms:
x2−4x+y2−6y=−9
Adding and subtracting the necessary constants to complete the square on
both xand yterms:
x2−4x+4+y2−6y+ 9 = −9+4+9
(x−2)2+ (y−3)2= 4
Step 2: Complete the square for the second equation by rearranging terms:
3x2−12x−2y2+ 8y= 19
Adding and subtracting the necessary constants to complete the square on
both xand yterms:
3(x2−4x)−2(y2−4y) = 19
3(x2−4x+ 4) −2(y2−4y+ 4) = 19 + 3(4) −2(4)
3(x−2)2−2(y−2)2= 27
(x−2)2
(3
√27 )2−(y−2)2
(2
√27 )2= 1
Step 3: Comparing the equations with the standard forms of conic sections:
The first equation represents a circle with center at (2, 3) and radius 2. The
second equation represents a hyperbola with center at (2, 2), horizontal axis
length 2√27, and vertical axis length 2√27/√2.
Therefore, the system of equations represents a circle and a hyperbola.
Question 12
Question
Let C(x) = 3x2−4x−10 be the cost function and R(x) = 4x+ 15 be the
revenue function for a company. Find the break-even point(s) for this company,
if they exist.
Solution
Step 1: To find the break-even point, we need to set the cost equal to the revenue
and solve for x. This means we need to solve the equation C(x) = R(x).
Step 2: First, we substitute the given functions into the equation:
3x2−4x−10 = 4x+ 15
10
Step 3: Next, we simplify the equation by moving all terms to one side:
3x2−4x−10 −4x−15 = 0
Step 4: Combine like terms:
3x2−8x−25 = 0
Step 5: Now, we need to solve the quadratic equation 3x2−8x−25 = 0.
We can do this by factoring or using the quadratic formula.
Step 6: Factoring the quadratic equation gives us:
(3x+ 5)(x−5) = 0
Step 7: Setting each factor equal to zero gives us the solutions:
3x+ 5 = 0 or x−5 = 0
Step 8: Solving these equations gives us:
x=−5
3or x= 5
Step 9: Therefore, the break-even points for the company are x=−5
3and
x= 5.
Question 13
Question
Solve the equation 5x2+ 12xy + 8y2−14x−34y−19 = 0 for yin terms of x.
Solution
Step 1: Rewrite the given equation in a standard form of a conic section.
Step 2: Complete the square for the quadratic terms to find the standard
form.
Step 3: Factor the quadratic expression and solve for yin terms of x.
Step 1: Rewrite the equation in a standard form of a conic section. We have
the equation:
5x2+ 12xy + 8y2−14x−34y−19 = 0
Step 2: Complete the square for the quadratic terms. To complete the square
for the xand yterms, we need to group the xterms, yterms, and constant
terms separately. The quadratic terms can be rewritten as:
(5x2+ 12xy + 8y2)−14x−34y−19 = 0
11
Now, complete the square for the quadratic terms (5x2+ 12xy + 8y2). We
have:
5(x2+12
5xy)+8y2−14x−34y−19 = 0
Step 3: Factor the quadratic expression. We can rewrite the quadratic part
as a perfect square trinomial:
5(x+6
5y)2
−36
5y2−14x−34y−19 = 0
Factor out the negative sign and simplify:
5(x+6
5y)2
−36
5(y2+5
2y)−14x−19 = 0
5(x+6
5y)2
−36
5(y2+5
2y+25
4)−14x−19 = 0
5(x+6
5y)2
−36
5(y+5
2)2−14x−19 = 0
Now, we can rewrite the equation in the standard form:
5(x+6
5y)2
−36
5(y+5
2)2= 14x+ 19
Therefore, the equation in terms of yis
36
5(y+5
2)2= 5(x+6
5y)2−14x−19
Now, solve for yin terms of x:
y+5
2=±√5
36(x+6
5y)2−14x+ 19
36
y=−5
2±√5
36(x+6
5y)2−14x+ 19
36
Question 14
Question
Find the standard form equation of the ellipse that satisfies the given conditions:
The ellipse has a major axis along the y-axis, a minor axis of length 6, and passes
through the points (0,5) and (0,−5).
12
Solution
Step 1: Since the major axis is along the y-axis, the equation of the ellipse will
be of the form y2
a2+x2
b2= 1, where ais the length of the semi-major axis and b
is the length of the semi-minor axis.
Step 2: Since the ”major axis” is the vertical axis, the length of the major
axis is 2a, and therefore 2a= 12 (from the given minor axis length of 6).
Step 3: Solving for a, we get a= 6.
Step 4: The center of the ellipse lies on the y-axis, so the x-coordinate of the
center is 0. Let’s denote the y-coordinate of the center as k.
Step 5: Using the midpoint formula, we find the center of the ellipse:
k=5+(−5)
2= 0
So, the center of the ellipse is at (0,0).
Step 6: We can now write the equation of the ellipse as:
y2
36 +x2
9= 1
This is the standard form equation of the ellipse.
Question 15
Question
Solve the following system of equations:
{x2+y2= 25
2x−y= 1
Solution
Step 1: Solve the second equation for yin terms of x:
2x−y= 1 =⇒y= 2x−1
Step 2: Substitute y= 2x−1into the first equation:
x2+ (2x−1)2= 25
Step 3: Expand and simplify the equation:
x2+ 4x2−4x+ 1 = 25
5x2−4x−24 = 0
13
Step 4: Factor the quadratic equation:
(5x+ 6)(x−4) = 0
Step 5: Set each factor equal to zero and solve for x:
5x+ 6 = 0 =⇒x=−6
5
x−4 = 0 =⇒x= 4
Step 6: Substitute x=−6
5and x= 4 back into y= 2x−1to find the
corresponding values of y: For x=−6
5:
y= 2 (−6
5)−1 = −12
5−1 = −17
5
For x= 4:
y= 2(4) −1 = 8 −1 = 7
Step 7: Therefore, the solution to the system of equations is:
{x=−6
5, y =−17
5
x= 4, y = 7
Question 16
Question
Find the standard form equation of the ellipse that satisfies the following con-
ditions: Center at (−3,2), major axis along the y-axis with a length of 8, and
minor axis along the x-axis with a length of 6.
Solution
Step 1: The general form of the equation of an ellipse centered at (h, k)with
major axis of length 2aalong the x-axis and minor axis of length 2balong the
y-axis is given by:
(x−h)2
a2+(y−k)2
b2= 1
Step 2: Since the center of the ellipse is (−3,2), we have h=−3and k= 2.
Step 3: The major axis length is 8, so 2a= 8 which gives a= 4.
Step 4: The minor axis length is 6, so 2b= 6 which gives b= 3.
Step 5: Now, substitute the values of h,k,a, and binto the equation of the
ellipse:
(x+ 3)2
42+(y−2)2
32= 1
14
Step 6: Simplify the equation:
(x+ 3)2
16 +(y−2)2
9= 1
Therefore, the standard form equation of the ellipse is (x+ 3)2
16 +(y−2)2
9= 1 .
Question 17
Question
Solve the system of equations:
{x2−y2= 9
2x+ 2y= 0
Solution
Step 1: Rearrange the second equation to solve for yin terms of x.
2x+ 2y= 0 ⇒2y=−2x⇒y=−x
Step 2: Substitute y=−xinto the first equation and solve for x.
x2−(−x)2= 9 ⇒x2−x2= 9 ⇒0 = 9
Step 3: Since 0= 9, the system of equations has no solution.
Question 18
Question
Solve the following system of equations:
{2x2+ 5y2−4x−20y−1 = 0
3x2−y2+ 2x−4y+ 2 = 0
Solution
Step 1: We will begin by completing the square for both equations. For the first
equation, we rewrite it as:
2(x2−2x) + 5(y2−4y) = 1
We complete the square for both xand yterms separately:
2(x2−2x+ 1) + 5(y2−4y+ 4) = 1 + 2 + 20
15
Simplifying gives:
2(x−1)2+ 5(y−2)2= 23
Step 2: Following the same process, we rewrite the second equation as:
3(x2+2
3x)−(y2−4y) = −2
Completing the square for both xand yterms separately:
3(x2+2
3x+(1
3)2)−(y2−4y+ 4) = −2+1+4
Simplifying gives:
3(x+1
3)2
−(y−2)2= 3
Step 3: Now we have the system in standard form:
{2(x−1)2+ 5(y−2)2= 23
3(x+1
3)2−(y−2)2= 3
Step 4: This system represents the equations of an ellipse and a hyperbola.
The solution to the system is the intersection points of these two conics. By
substituting the second equation into the first, or vice versa, we can solve for x
and y.
Unfortunately, the algebra required to solve this system of equations exceeds
the complexity of a typical College Algebra course.
Question 19
Question
Solve the system of equations:
{3x2−y2= 4
x+y= 2
Solution
Step 1: Let’s first solve the second equation for xin terms of y.
x+y= 2
x= 2 −y
16
Step 2: Substitute x= 2 −yinto the first equation.
3(2 −y)2−y2= 4
3(4 −4y+y2)−y2= 4
12 −12y+ 3y2−y2= 4
12 −12y+ 2y2= 4
2y2−12y+ 8 = 0
y2−6y+ 4 = 0
Step 3: Solve the quadratic equation y2−6y+ 4 = 0 using the quadratic
formula.
y=−(−6) ±√(−6)2−4(1)(4)
2(1)
y=6±√36 −16
2
y=6±√20
2
y= 3 ±2√5
Step 4: Find the corresponding values of xfor each solution of y.
For y= 3 + 2√5 :
x= 2 −(3 + 2√5) = −1−2√5
For y= 3 −2√5 :
x= 2 −(3 −2√5) = −1+2√5
Step 5: Therefore, the solutions to the system of equations are:
{x=−1−2√5, y = 3 + 2√5
x=−1+2√5, y = 3 −2√5
Question 20
Question
Solve the system of equations:
{x2−y2= 4
xy = 1
Solution
Step 1: Let’s solve for yin the second equation xy = 1:
y=1
x
17
Step 2: Substitute y=1
xinto the first equation x2−y2= 4:
x2−(1
x)2
= 4
Step 3: Simplify the equation:
x2−1
x2= 4
Step 4: Multiply through by x2to clear the fraction:
x4−1 = 4x2
Step 5: Rearrange the equation:
x4−4x2−1 = 0
Step 6: This is a quadratic equation in terms of x2. Let u=x2:
u2−4u−1 = 0
Step 7: Solve this quadratic equation for uusing the quadratic formula:
u=4±√42−4(1)(−1)
2(1)
u=4±√16 + 4
2
u=4±√20
2
u=4±2√5
2
u= 2 ±√5
Step 8: Since u=x2, there are two possible solutions for x:x2= 2 + √5
and x2= 2 −√5.
Step 9: Solve for x: For x2= 2 + √5,x=√2 + √5or x=−√2 + √5. For
x2= 2 −√5,x=√2−√5or x=−√2−√5.
Step 10: Substituting the values of xback into y=1
xto find the corre-
sponding values of y: For x=√2 + √5,y=1
√2+√5or y=−1
√2+√5. For
x=−√2 + √5,y=−1
√2+√5or y=1
√2+√5. Similarly for the other two
values of x.
Therefore, the solutions to the system of equations are:
(x, y) = (√2 + √5,1
√2 + √5),(−√2 + √5,−1
√2 + √5),(√2−√5,1
√2−√5),(−√2−√5,−1
√2−√5)
18
Question 21
Question
Given the equation of a parabolic reflector: y=x2−4, find the focus and
directrix.
Solution
Step 1: To find the vertex of the parabolic reflector, rewrite the equation in
vertex form. The vertex form of a parabolic equation is y=a(x−h)2+k,
where (h, k)is the vertex. Step 2: Compare the given equation with the vertex
form and identify hand k. Step 3: The vertex of the parabolic reflector is at
(h, k). Step 4: The focus of a parabolic reflector is given by the point (h, k+1
4a).
Step 5: The directrix of a parabolic reflector is the line y=k−1
4a.
Therefore, the focus and directrix of the parabolic reflector y=x2−4are
found by substituting a= 1,h= 0, and k=−4into the formulas for the focus
and directrix.
Question 22
Question
Let y=−3x2+ 4x−2be the equation of a parabola. Find the coordinates of
the vertex, the focus, and the directrix of this parabola.
Solution
Step 1: To find the vertex of the parabola, use the formula x=−b
2ato find the
x-coordinate of the vertex. Then substitute this x value back into the equation
to find the y-coordinate.
Step 1: Find the x-coordinate of the vertex.
The quadratic equation is in the form y=ax2+bx +c, so a=−3,b= 4.
Substitute these values into x=−b
2a.
x=−4
2(−3) =2
3
Step 2: Find the y-coordinate of the vertex.
Plug x=2
3back into the equation y=−3x2+ 4x−2.
y=−3(2
3)2
+ 4 (2
3)−2
y=−2
19
Thus, the vertex of the parabola is (2
3,−2).
Step 2: To find the focus of the parabola, use the formula p=1
4awhere pis
the distance from the vertex to the focus.
Step 3: Find the distance from the vertex to the focus.
p=1
4(−3) =−1
12 (Note: The negative sign indicates that the parabola opens
downwards)
Step 4: Find the focus.
Since the parabola opens downwards, the focus is at a distance pbelow the
vertex. The y-coordinate of the focus is −2−1
12 =−25
12 .
Therefore, the focus of the parabola is (2
3,−25
12 ).
Step 3: To find the equation of the directrix, use the formula y=k+1
4a
where kis the y-coordinate of the vertex.
Step 5: Find the equation of the directrix.
y=−2 + 1
4(−3) =−2−1
12 =−25
12
Hence, the equation of the directrix is y=−25
12 .
Question 23
Question
Solve the following system of equations:
{x2+y2= 25
x−y= 1
Solution
Step 1: Solve the second equation for xin terms of y:
x=y+ 1
Step 2: Substitute x=y+ 1 into the first equation:
(y+ 1)2+y2= 25
Step 3: Expand the left side of the equation:
y2+ 2y+1+y2= 25
Step 4: Combine like terms:
2y2+ 2y−24 = 0
20
Step 5: Divide the equation by 2 to simplify it,
y2+y−12 = 0
Step 6: Factor the quadratic equation:
(y+ 4)(y−3) = 0
Step 7: Solve for y,
y=−4,3
Step 8: Substitute the values of yback into x=y+1 to find the correspond-
ing values of x:
x=−4 + 1 = −3, x = 3 + 1 = 4
Step 9: The solutions to the system of equations are (−3,−4) and (4,3).
Question 24
Question
Find the standard form of the equation of the hyperbola with foci at (−5,0)
and (5,0) and vertices at (−7,0) and (7,0).
Solution
Step 1: Identify the center of the hyperbola by finding the midpoint of the
segment connecting the vertices. The center is the point (h, k).
Midpoint =(−7+7
2,0+0
2)= (0,0)
Step 2: Determine the value of aby calculating the distance from the center
to one of the vertices. In this case, a= 7.
Step 3: Find the value of c, which is the distance from the center to one of
the foci.
c= 5
Step 4: Use the relationship c2=a2+b2to solve for b.
25 = 49 + b2
b2= 25 −49 = −24
Step 5: Write the equation of the hyperbola in standard form (x−h)2
a2−
(y−k)2
b2= 1.
x2
49 −y2
−24 = 1
Therefore, the standard form of the equation of the hyperbola is x2
49 +y2
24 = 1.
21
Question 25
Question
Find the standard form of the equation of a circle passing through the points
(1,2) and (5,6).
Solution
Step 1: Use the general equation of a circle in standard form: (x−h)2+(y−k)2=
r2, where (h, k)is the center of the circle and ris the radius.
Step 2: Find the midpoint of the given points to get the center of the circle
(h, k). Midpoint formula: (x1+x2
2,y1+y2
2).
Plugging in the points (1,2) and (5,6):h=1+5
2= 3, and k=2+6
2= 4, so
the center is at (3,4).
Step 3: Calculate the radius rby substituting one of the given points into the
equation of the circle and solving for r. We will use (1,2).(1−3)2+(2−4)2=r2
(−2)2+ (−2)2=r24 + 4 = r2r2= 8 r=√8 = 2√2.
Step 4: Substitute the center and radius into the standard form of the equa-
tion of a circle. (x−3)2+ (y−4)2= (2√2)2(x−3)2+ (y−4)2= 8
Therefore, the standard form of the equation of the circle passing through
the points (1,2) and (5,6) is (x−3)2+ (y−4)2= 8.
22
b2= 11
b=√11.
Step 5: Write the equation of the hyperbola in standard form (x2
a2−y2
b2= 1).
The equation of the hyperbola is therefore:
x2
9−y2
11 = 1.
Question 2
Question
Find the standard form equation of a parabola with a focus at (−3,2) and a
directrix given by y=−4.
Solution
Step 1: Recall that the standard form equation of a parabola with a horizontal
axis of symmetry and a focus at (h+k, k)is given by
(x−h)2= 4p(y−k),
where pis the distance from the vertex to the focus.
Step 2: In this problem, we know that the focus is at (−3,2), so h=−3and
k= 2.
Step 3: The given directrix is y=−4, which means the distance from the
vertex to the directrix is |k−(−4)|=|2+4|= 6.
Step 4: Since the parabola opens upwards from the directrix, p= 6.
Step 5: Now, we can plug in the values of h,k, and pinto the standard form
equation to find the equation of the parabola.
(x+ 3)2= 4 ·6(y−2)
Step 6: Simplify the equation to get the standard form of the parabola.
(x+ 3)2= 24(y−2)
Step 7: Therefore, the standard form equation of the parabola with a focus
at (−3,2) and a directrix given by y=−4is (x+ 3)2= 24(y−2).
Question 3
Question
Given the equation of a hyperbola as 4x2−9y2= 36, find the center, vertices,
foci, and equations of the asymptotes.
2
Solution
Step 1: Rewrite the given equation in standard form by dividing each term on
both sides by 36.
x2
9−y2
4= 1
Step 2: Identify the center of the hyperbola, which is at the point (h, k) =
(0,0).
Step 3: Determine the vertices of the hyperbola. Since the hyperbola opens
horizontally, the vertices are located at (h±a, k), where a= 3. So, the vertices
are at (−3,0) and (3,0).
Step 4: Find the foci of the hyperbola. The distance from the center to each
focus is given by c=√a2+b2. Substitute a= 3 and b= 2 into the formula to
find c.
c=√32+ 22=√13
Therefore, the foci are located at (√13,0) and (−√13,0).
Step 5: Calculate the equations of the asymptotes. The general form for
the asymptotes is y=±b
ax. Substitute a= 3 and b= 2 into the equation to
find the equations of the asymptotes. So, the equations of the asymptotes are
y=±2
3x.
Question 4
Question
Solve the following system of equations:
{2x2−3y2= 1
3x2−2y2=−1
Solution
Step 1: Rewrite the system of equations in standard form by eliminating y2.
{2x2−3y2= 1
3x2−2y2=−1
To eliminate y2, we will multiply the first equation by 2 and the second
equation by 3:
{4x2−6y2= 2
9x2−6y2=−3
Step 2: Subtract the second equation from the first equation to eliminate
y2.
(4x2−6y2)−(9x2−6y2) = 2 −(−3)
3
Simplify to get:
−5x2= 5
Step 3: Solve for x. Divide both sides by −5to solve for x:
x2=−1
Since x2=−1has no real solutions, the system of equations has no real solu-
tions.
Question 5
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: Determine the center of the hyperbola. Since the center of a hyperbola
is the midpoint of the line segment connecting the vertices, the center of the
hyperbola is (0,0).
Step 2: Find the distance between the center and a vertex to determine a.
The distance between the center (0,0) and either of the vertices (3,0) or
(−3,0) is 3 units.
So a= 3.
Step 3: Find the distance between the center and a focus to determine c.
The distance between the center (0,0) and either of the foci (5,0) or (−5,0)
is 5 units.
So c= 5.
Step 4: Use the relationship c2=a2+b2to find b.
Since we have a= 3 and c= 5, we can substitute these values into the
equation to solve for b:
52= 32+b2
25 = 9 + b2
b2= 16
b=±4
Step 5: Write the equation of the hyperbola in standard form.
The equation of a hyperbola centered at the origin with vertices on the x-axis
is x2
a2−y2
b2= 1.
Since a= 3,b= 4, the equation of the hyperbola is:
x2
9−y2
16 = 1
4
Question 6
Question
Find the standard form equation of a hyperbola with vertices at (−2,1) and
(4,1), and the distance between its foci is √29.
Solution
Step 1: First, we find the center of the hyperbola by finding the midpoint of the
vertices. The midpoint formula is given by
(x1+x2
2,y1+y2
2).
Step 2: Applying the midpoint formula, we find the center of the hyperbola:
(−2+4
2,1+1
2)= (1,1).
Step 3: Since the distance between the vertices is 6units, we can determine
a, the distance from the center to a vertex, which is 3units.
Step 4: Given that the distance between the foci is √29, and that for a
hyperbola c2=a2+b2, where cis the distance between the foci, we have
c2= 29. Thus, c=√29.
Step 5: We know that c2=a2+b2, where a= 3. Plugging in the values, we
get
29 = 9 + b2,
b2= 29 −9 = 20.
Step 6: Now, we can write the standard form equation of the hyperbola
centered at (1,1):
(x−1)2
9−(y−1)2
20 = 1.
Therefore, the standard form equation of the hyperbola is (x−1)2
9−(y−1)2
20 =
1.
Question 7
Question
Solve the equation 3x2−12x−9y2+ 36y= 0 for yin terms of x.
5
Solution
Step 1: Rearrange the equation to isolate the terms involving y:
3x2−12x= 9y2−36y
Step 2: Factor out the coefficients in front of y2and y:
3(x2−4x) = 9(y2−4y)
Step 3: Complete the square on the right side of the equation by adding
(4/2)2= 4 to both sides:
3(x2−4x+ 4) = 9(y2−4y+ 4)
Step 4: Simplify both sides of the equation:
3(x−2)2= 9(y−2)2
Step 5: Divide both sides by 9 to solve for y:
(x−2)2= 3(y−2)2
Step 6: Take the square root of both sides:
√(x−2)2=±√3(y−2)2
Step 7: Simplify:
|x−2|=√3|y−2|
Step 8: Solve for yby isolating it on one side of the equation:
|y−2|=1
√3|x−2|
Therefore, the equation in terms of yis |y−2|=1
√3|x−2|.
Question 8
Question
Solve the system of equations:
{x2+y2= 25
xy = 12
6
Solution
Step 1: Let’s solve the second equation for yin terms of x.
xy = 12 =⇒y=12
x
Step 2: Substitute y=12
xinto the first equation.
x2+(12
x)2
= 25
Step 3: Simplify the equation.
x2+144
x2= 25
Step 4: Multiply through by x2to clear the fraction.
x4+ 144 = 25x2
Step 5: Rearrange the equation to form a quadratic equation.
x4−25x2+ 144 = 0
Step 6: Letting u=x2, the equation becomes
u2−25u+ 144 = 0
Step 7: Solve the quadratic equation by factoring or using the quadratic
formula. The factors of 144 that add up to -25 are -16 and -9. So, the equation
factors as (u−16)(u−9) = 0.
Step 8: Solve for u.
u= 16 or u= 9
Step 9: Substitute u=x2back in.
x2= 16 or x2= 9
Step 10: Solve for x.
x=±4or x=±3
Step 11: Substitute xinto y=12
xto find the corresponding values of y. For
x= 4,y=12
4= 3 or y=−3. For x=−4,y=−3or y= 3. For x= 3,
y=12
3= 4 or y=−4. For x=−3,y=−4or y= 4.
Step 12: Therefore, the solutions to the system are (−4,−3),(−4,3),(4,−3),
and (4,3).
7
Question 9
Question
Solve the following system of equations:
{x2+y2= 25
x−y= 3
Solution
Step 1: Rewrite the second equation as 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 for yusing the quadratic formula:
y=−6±√62−4(2)(−16)
2(2)
y=−6±√36 + 128
4
y=−6±√164
4
y=−6±2√41
4
y=−3±√41
2
Step 5: Substitute the values of yback into the equation x=y+ 3 to find
the corresponding values of x:
x=−3 + √41
2+ 3 or x=−3−√41
2+ 3
x=−3 + √41 + 6
2or x=−3−√41 + 6
2
x=3 + √41
2or x=3−√41
2
Step 6: Therefore, the solutions to the system of equations are:
(x, y) = (3 + √41
2,−3 + √41
2)and (3−√41
2,−3−√41
2)
8
Question 10
Question
Solve the equation 4x2+ 9y2−24x+ 54y−45 = 0 and determine the type of
conic section represented by this equation.
Solution
Step 1: Rearrange the equation by completing the square for both xand y
terms.
4x2+ 9y2−24x+ 54y−45 = 0
4(x2−6x) + 9(y2+ 6y) = 45
4(x2−6x+ 9) + 9(y2+ 6y+ 9) = 45 + 36 + 81
4(x−3)2+ 9(y+ 3)2= 162
Step 2: Divide by 162 to normalize the equation.
4(x−3)2
162 +9(y+ 3)2
162 = 1
(x−3)2
162
4
+(y+ 3)2
162
9
= 1
(x−3)2
40.5+(y+ 3)2
18 = 1
Step 3: Compare the equation with the standard form of conic sections to
determine the type of conic section. The given equation matches the standard
form of an ellipse:
((x−h)2
a2)+((y−k)2
b2) = 1
where the center C(h, k), major axis length 2a, and minor axis length 2b. In
this case, C(3,−3),a=√40.5, and b=√18, so the conic section represented
by the equation is an ellipse.
Question 11
Question
Solve the following system of equations and identify the type of conic section
represented by the graph of the equations: {x2+y2−4x−6y+ 9 = 0
3x2−2y2−12x+ 8y−19 = 0
9
Solution
Step 1: Complete the square for the first equation by rearranging terms:
x2−4x+y2−6y=−9
Adding and subtracting the necessary constants to complete the square on
both xand yterms:
x2−4x+4+y2−6y+ 9 = −9+4+9
(x−2)2+ (y−3)2= 4
Step 2: Complete the square for the second equation by rearranging terms:
3x2−12x−2y2+ 8y= 19
Adding and subtracting the necessary constants to complete the square on
both xand yterms:
3(x2−4x)−2(y2−4y) = 19
3(x2−4x+ 4) −2(y2−4y+ 4) = 19 + 3(4) −2(4)
3(x−2)2−2(y−2)2= 27
(x−2)2
(3
√27 )2−(y−2)2
(2
√27 )2= 1
Step 3: Comparing the equations with the standard forms of conic sections:
The first equation represents a circle with center at (2, 3) and radius 2. The
second equation represents a hyperbola with center at (2, 2), horizontal axis
length 2√27, and vertical axis length 2√27/√2.
Therefore, the system of equations represents a circle and a hyperbola.
Question 12
Question
Let C(x) = 3x2−4x−10 be the cost function and R(x) = 4x+ 15 be the
revenue function for a company. Find the break-even point(s) for this company,
if they exist.
Solution
Step 1: To find the break-even point, we need to set the cost equal to the revenue
and solve for x. This means we need to solve the equation C(x) = R(x).
Step 2: First, we substitute the given functions into the equation:
3x2−4x−10 = 4x+ 15
10
Step 3: Next, we simplify the equation by moving all terms to one side:
3x2−4x−10 −4x−15 = 0
Step 4: Combine like terms:
3x2−8x−25 = 0
Step 5: Now, we need to solve the quadratic equation 3x2−8x−25 = 0.
We can do this by factoring or using the quadratic formula.
Step 6: Factoring the quadratic equation gives us:
(3x+ 5)(x−5) = 0
Step 7: Setting each factor equal to zero gives us the solutions:
3x+ 5 = 0 or x−5 = 0
Step 8: Solving these equations gives us:
x=−5
3or x= 5
Step 9: Therefore, the break-even points for the company are x=−5
3and
x= 5.
Question 13
Question
Solve the equation 5x2+ 12xy + 8y2−14x−34y−19 = 0 for yin terms of x.
Solution
Step 1: Rewrite the given equation in a standard form of a conic section.
Step 2: Complete the square for the quadratic terms to find the standard
form.
Step 3: Factor the quadratic expression and solve for yin terms of x.
Step 1: Rewrite the equation in a standard form of a conic section. We have
the equation:
5x2+ 12xy + 8y2−14x−34y−19 = 0
Step 2: Complete the square for the quadratic terms. To complete the square
for the xand yterms, we need to group the xterms, yterms, and constant
terms separately. The quadratic terms can be rewritten as:
(5x2+ 12xy + 8y2)−14x−34y−19 = 0
11
Now, complete the square for the quadratic terms (5x2+ 12xy + 8y2). We
have:
5(x2+12
5xy)+8y2−14x−34y−19 = 0
Step 3: Factor the quadratic expression. We can rewrite the quadratic part
as a perfect square trinomial:
5(x+6
5y)2
−36
5y2−14x−34y−19 = 0
Factor out the negative sign and simplify:
5(x+6
5y)2
−36
5(y2+5
2y)−14x−19 = 0
5(x+6
5y)2
−36
5(y2+5
2y+25
4)−14x−19 = 0
5(x+6
5y)2
−36
5(y+5
2)2−14x−19 = 0
Now, we can rewrite the equation in the standard form:
5(x+6
5y)2
−36
5(y+5
2)2= 14x+ 19
Therefore, the equation in terms of yis
36
5(y+5
2)2= 5(x+6
5y)2−14x−19
Now, solve for yin terms of x:
y+5
2=±√5
36(x+6
5y)2−14x+ 19
36
y=−5
2±√5
36(x+6
5y)2−14x+ 19
36
Question 14
Question
Find the standard form equation of the ellipse that satisfies the given conditions:
The ellipse has a major axis along the y-axis, a minor axis of length 6, and passes
through the points (0,5) and (0,−5).
12
Solution
Step 1: Since the major axis is along the y-axis, the equation of the ellipse will
be of the form y2
a2+x2
b2= 1, where ais the length of the semi-major axis and b
is the length of the semi-minor axis.
Step 2: Since the ”major axis” is the vertical axis, the length of the major
axis is 2a, and therefore 2a= 12 (from the given minor axis length of 6).
Step 3: Solving for a, we get a= 6.
Step 4: The center of the ellipse lies on the y-axis, so the x-coordinate of the
center is 0. Let’s denote the y-coordinate of the center as k.
Step 5: Using the midpoint formula, we find the center of the ellipse:
k=5+(−5)
2= 0
So, the center of the ellipse is at (0,0).
Step 6: We can now write the equation of the ellipse as:
y2
36 +x2
9= 1
This is the standard form equation of the ellipse.
Question 15
Question
Solve the following system of equations:
{x2+y2= 25
2x−y= 1
Solution
Step 1: Solve the second equation for yin terms of x:
2x−y= 1 =⇒y= 2x−1
Step 2: Substitute y= 2x−1into the first equation:
x2+ (2x−1)2= 25
Step 3: Expand and simplify the equation:
x2+ 4x2−4x+ 1 = 25
5x2−4x−24 = 0
13
Step 4: Factor the quadratic equation:
(5x+ 6)(x−4) = 0
Step 5: Set each factor equal to zero and solve for x:
5x+ 6 = 0 =⇒x=−6
5
x−4 = 0 =⇒x= 4
Step 6: Substitute x=−6
5and x= 4 back into y= 2x−1to find the
corresponding values of y: For x=−6
5:
y= 2 (−6
5)−1 = −12
5−1 = −17
5
For x= 4:
y= 2(4) −1 = 8 −1 = 7
Step 7: Therefore, the solution to the system of equations is:
{x=−6
5, y =−17
5
x= 4, y = 7
Question 16
Question
Find the standard form equation of the ellipse that satisfies the following con-
ditions: Center at (−3,2), major axis along the y-axis with a length of 8, and
minor axis along the x-axis with a length of 6.
Solution
Step 1: The general form of the equation of an ellipse centered at (h, k)with
major axis of length 2aalong the x-axis and minor axis of length 2balong the
y-axis is given by:
(x−h)2
a2+(y−k)2
b2= 1
Step 2: Since the center of the ellipse is (−3,2), we have h=−3and k= 2.
Step 3: The major axis length is 8, so 2a= 8 which gives a= 4.
Step 4: The minor axis length is 6, so 2b= 6 which gives b= 3.
Step 5: Now, substitute the values of h,k,a, and binto the equation of the
ellipse:
(x+ 3)2
42+(y−2)2
32= 1
14
Step 6: Simplify the equation:
(x+ 3)2
16 +(y−2)2
9= 1
Therefore, the standard form equation of the ellipse is (x+ 3)2
16 +(y−2)2
9= 1 .
Question 17
Question
Solve the system of equations:
{x2−y2= 9
2x+ 2y= 0
Solution
Step 1: Rearrange the second equation to solve for yin terms of x.
2x+ 2y= 0 ⇒2y=−2x⇒y=−x
Step 2: Substitute y=−xinto the first equation and solve for x.
x2−(−x)2= 9 ⇒x2−x2= 9 ⇒0 = 9
Step 3: Since 0= 9, the system of equations has no solution.
Question 18
Question
Solve the following system of equations:
{2x2+ 5y2−4x−20y−1 = 0
3x2−y2+ 2x−4y+ 2 = 0
Solution
Step 1: We will begin by completing the square for both equations. For the first
equation, we rewrite it as:
2(x2−2x) + 5(y2−4y) = 1
We complete the square for both xand yterms separately:
2(x2−2x+ 1) + 5(y2−4y+ 4) = 1 + 2 + 20
15
Simplifying gives:
2(x−1)2+ 5(y−2)2= 23
Step 2: Following the same process, we rewrite the second equation as:
3(x2+2
3x)−(y2−4y) = −2
Completing the square for both xand yterms separately:
3(x2+2
3x+(1
3)2)−(y2−4y+ 4) = −2+1+4
Simplifying gives:
3(x+1
3)2
−(y−2)2= 3
Step 3: Now we have the system in standard form:
{2(x−1)2+ 5(y−2)2= 23
3(x+1
3)2−(y−2)2= 3
Step 4: This system represents the equations of an ellipse and a hyperbola.
The solution to the system is the intersection points of these two conics. By
substituting the second equation into the first, or vice versa, we can solve for x
and y.
Unfortunately, the algebra required to solve this system of equations exceeds
the complexity of a typical College Algebra course.
Question 19
Question
Solve the system of equations:
{3x2−y2= 4
x+y= 2
Solution
Step 1: Let’s first solve the second equation for xin terms of y.
x+y= 2
x= 2 −y
16
Step 2: Substitute x= 2 −yinto the first equation.
3(2 −y)2−y2= 4
3(4 −4y+y2)−y2= 4
12 −12y+ 3y2−y2= 4
12 −12y+ 2y2= 4
2y2−12y+ 8 = 0
y2−6y+ 4 = 0
Step 3: Solve the quadratic equation y2−6y+ 4 = 0 using the quadratic
formula.
y=−(−6) ±√(−6)2−4(1)(4)
2(1)
y=6±√36 −16
2
y=6±√20
2
y= 3 ±2√5
Step 4: Find the corresponding values of xfor each solution of y.
For y= 3 + 2√5 :
x= 2 −(3 + 2√5) = −1−2√5
For y= 3 −2√5 :
x= 2 −(3 −2√5) = −1+2√5
Step 5: Therefore, the solutions to the system of equations are:
{x=−1−2√5, y = 3 + 2√5
x=−1+2√5, y = 3 −2√5
Question 20
Question
Solve the system of equations:
{x2−y2= 4
xy = 1
Solution
Step 1: Let’s solve for yin the second equation xy = 1:
y=1
x
17
Step 2: Substitute y=1
xinto the first equation x2−y2= 4:
x2−(1
x)2
= 4
Step 3: Simplify the equation:
x2−1
x2= 4
Step 4: Multiply through by x2to clear the fraction:
x4−1 = 4x2
Step 5: Rearrange the equation:
x4−4x2−1 = 0
Step 6: This is a quadratic equation in terms of x2. Let u=x2:
u2−4u−1 = 0
Step 7: Solve this quadratic equation for uusing the quadratic formula:
u=4±√42−4(1)(−1)
2(1)
u=4±√16 + 4
2
u=4±√20
2
u=4±2√5
2
u= 2 ±√5
Step 8: Since u=x2, there are two possible solutions for x:x2= 2 + √5
and x2= 2 −√5.
Step 9: Solve for x: For x2= 2 + √5,x=√2 + √5or x=−√2 + √5. For
x2= 2 −√5,x=√2−√5or x=−√2−√5.
Step 10: Substituting the values of xback into y=1
xto find the corre-
sponding values of y: For x=√2 + √5,y=1
√2+√5or y=−1
√2+√5. For
x=−√2 + √5,y=−1
√2+√5or y=1
√2+√5. Similarly for the other two
values of x.
Therefore, the solutions to the system of equations are:
(x, y) = (√2 + √5,1
√2 + √5),(−√2 + √5,−1
√2 + √5),(√2−√5,1
√2−√5),(−√2−√5,−1
√2−√5)
18
Question 21
Question
Given the equation of a parabolic reflector: y=x2−4, find the focus and
directrix.
Solution
Step 1: To find the vertex of the parabolic reflector, rewrite the equation in
vertex form. The vertex form of a parabolic equation is y=a(x−h)2+k,
where (h, k)is the vertex. Step 2: Compare the given equation with the vertex
form and identify hand k. Step 3: The vertex of the parabolic reflector is at
(h, k). Step 4: The focus of a parabolic reflector is given by the point (h, k+1
4a).
Step 5: The directrix of a parabolic reflector is the line y=k−1
4a.
Therefore, the focus and directrix of the parabolic reflector y=x2−4are
found by substituting a= 1,h= 0, and k=−4into the formulas for the focus
and directrix.
Question 22
Question
Let y=−3x2+ 4x−2be the equation of a parabola. Find the coordinates of
the vertex, the focus, and the directrix of this parabola.
Solution
Step 1: To find the vertex of the parabola, use the formula x=−b
2ato find the
x-coordinate of the vertex. Then substitute this x value back into the equation
to find the y-coordinate.
Step 1: Find the x-coordinate of the vertex.
The quadratic equation is in the form y=ax2+bx +c, so a=−3,b= 4.
Substitute these values into x=−b
2a.
x=−4
2(−3) =2
3
Step 2: Find the y-coordinate of the vertex.
Plug x=2
3back into the equation y=−3x2+ 4x−2.
y=−3(2
3)2
+ 4 (2
3)−2
y=−2
19
Thus, the vertex of the parabola is (2
3,−2).
Step 2: To find the focus of the parabola, use the formula p=1
4awhere pis
the distance from the vertex to the focus.
Step 3: Find the distance from the vertex to the focus.
p=1
4(−3) =−1
12 (Note: The negative sign indicates that the parabola opens
downwards)
Step 4: Find the focus.
Since the parabola opens downwards, the focus is at a distance pbelow the
vertex. The y-coordinate of the focus is −2−1
12 =−25
12 .
Therefore, the focus of the parabola is (2
3,−25
12 ).
Step 3: To find the equation of the directrix, use the formula y=k+1
4a
where kis the y-coordinate of the vertex.
Step 5: Find the equation of the directrix.
y=−2 + 1
4(−3) =−2−1
12 =−25
12
Hence, the equation of the directrix is y=−25
12 .
Question 23
Question
Solve the following system of equations:
{x2+y2= 25
x−y= 1
Solution
Step 1: Solve the second equation for xin terms of y:
x=y+ 1
Step 2: Substitute x=y+ 1 into the first equation:
(y+ 1)2+y2= 25
Step 3: Expand the left side of the equation:
y2+ 2y+1+y2= 25
Step 4: Combine like terms:
2y2+ 2y−24 = 0
20
Step 5: Divide the equation by 2 to simplify it,
y2+y−12 = 0
Step 6: Factor the quadratic equation:
(y+ 4)(y−3) = 0
Step 7: Solve for y,
y=−4,3
Step 8: Substitute the values of yback into x=y+1 to find the correspond-
ing values of x:
x=−4 + 1 = −3, x = 3 + 1 = 4
Step 9: The solutions to the system of equations are (−3,−4) and (4,3).
Question 24
Question
Find the standard form of the equation of the hyperbola with foci at (−5,0)
and (5,0) and vertices at (−7,0) and (7,0).
Solution
Step 1: Identify the center of the hyperbola by finding the midpoint of the
segment connecting the vertices. The center is the point (h, k).
Midpoint =(−7+7
2,0+0
2)= (0,0)
Step 2: Determine the value of aby calculating the distance from the center
to one of the vertices. In this case, a= 7.
Step 3: Find the value of c, which is the distance from the center to one of
the foci.
c= 5
Step 4: Use the relationship c2=a2+b2to solve for b.
25 = 49 + b2
b2= 25 −49 = −24
Step 5: Write the equation of the hyperbola in standard form (x−h)2
a2−
(y−k)2
b2= 1.
x2
49 −y2
−24 = 1
Therefore, the standard form of the equation of the hyperbola is x2
49 +y2
24 = 1.
21
Question 25
Question
Find the standard form of the equation of a circle passing through the points
(1,2) and (5,6).
Solution
Step 1: Use the general equation of a circle in standard form: (x−h)2+(y−k)2=
r2, where (h, k)is the center of the circle and ris the radius.
Step 2: Find the midpoint of the given points to get the center of the circle
(h, k). Midpoint formula: (x1+x2
2,y1+y2
2).
Plugging in the points (1,2) and (5,6):h=1+5
2= 3, and k=2+6
2= 4, so
the center is at (3,4).
Step 3: Calculate the radius rby substituting one of the given points into the
equation of the circle and solving for r. We will use (1,2).(1−3)2+(2−4)2=r2
(−2)2+ (−2)2=r24 + 4 = r2r2= 8 r=√8 = 2√2.
Step 4: Substitute the center and radius into the standard form of the equa-
tion of a circle. (x−3)2+ (y−4)2= (2√2)2(x−3)2+ (y−4)2= 8
Therefore, the standard form of the equation of the circle passing through
the points (1,2) and (5,6) is (x−3)2+ (y−4)2= 8.
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