MATH 121 - COLLEGE ALGEBRA -
Equations and properties of conic
sections
Question Bank - Set 1
Liberty University
Question 1
Question
Find the standard form of the equation of the conic section represented by the
equation 9x2−16y2+ 54x−64y= 169.
Solution
To find the standard form of the equation for the given conic section, we need
to complete the square for both the xand yterms.
Step 1: Grouping the xand yterms We start by grouping the xterms
together and the yterms together:
9x2+ 54x−16y2−64y= 169
Step 2: Completing the square for xFor the xterms, we complete the
square by adding and subtracting (54/2)2= 729 inside the parentheses:
9(x2+ 6x+ 9) −16y2−64y= 169 + 9(9)
Step 3: Completing the square for yFor the yterms, we complete the
square by adding and subtracting (64/2)2= 1024 inside the parentheses:
9(x2+ 6x+ 9) −16(y2+ 4y+ 4) = 169 + 81 + 256
Step 4: Rewrite the equation Now we can rewrite the equation in the
standard form by factoring the completed square terms:
9(x+ 3)2−16(y+ 2)2= 506
Thus, the standard form of the equation for the conic section represented by
9x2−16y2+ 54x−64y= 169 is 9(x+ 3)2−16(y+ 2)2= 506.
Question 2
Question
Find the standard form of the equation of the parabola with focus F(0,3) and
directrix y=−3.
Solution
Step 1: Recall the definition of a parabola. A parabola is the set of all points
that are equidistant from a fixed point called the focus and a fixed line called
the directrix.
Step 2: Given that the focus is F(0,3) and the directrix is y=−3, we can
see that the vertex is the midpoint between the focus and the directrix. So, the
vertex is at (0,0).
Step 3: Since the directrix is a horizontal line, we have a vertical parabola
opening either upwards or downwards. Since the focus is above the directrix,
the parabola opens upwards.
Step 4: The distance between the vertex and the focus is the same as the
distance between the vertex and the directrix. This distance is called the focal
length. In this case, the focal length is 3units.
Step 5: The standard form of the equation of a parabola with vertex at (h, k)
and focal length |p|is given by:
(x−h)2= 4p(y−k)
Step 6: Plugging in the values h= 0,k= 0, and p= 3 into the standard
form equation, we get:
x2= 12y
Therefore, the standard form of the equation of the parabola is x2= 12y.
Question 3
Question
Solve the following system of equations:
{x2+y2= 25
x+ 2y= 10
Solution
Step 1: Solve the second equation for xin terms of y:
x= 10 −2y
2
Step 2: Substitute x= 10 −2yinto the first equation:
(10 −2y)2+y2= 25
Step 3: Expand and simplify the equation:
100 −40y+ 4y2+y2= 25
Step 4: Rearrange the equation into standard quadratic form:
5y2−40y+ 75 = 0
Step 5: Solve the quadratic equation for yusing the quadratic formula:
y=−(−40) ±√(−40)2−4(5)(75)
2(5)
Step 6: Simplify the expression under the square root:
y=40 ±√1600 −1500
10
y=40 ±√100
10
Step 7: Solve for y:
y=40 ±10
10
y= 5,3
Step 8: Substitute y= 5 into x= 10 −2y:
x= 10 −2(5) = 0
Step 9: Substitute y= 3 into x= 10 −2y:
x= 10 −2(3) = 4
Step 10: The solutions to the system are (0,5) and (4,3).
Question 4
Question
Solve the following system of equations:
{x2+y2= 25
x+y= 7
3
Solution
Step 1: Rewrite the second equation in terms of one variable. Solve for either
xor yin terms of the other variable. We will solve for yin terms of x.
y= 7 −x
Step 2: Substitute the expression for yinto the first equation.
x2+ (7 −x)2= 25
Step 3: Expand and simplify the equation.
x2+ 49 −14x+x2= 25
2x2−14x+ 24 = 0
Step 4: Rearrange the equation into standard quadratic form.
x2−7x+ 12 = 0
Step 5: Factor the quadratic equation.
(x−3)(x−4) = 0
Step 6: Set each factor to zero and solve for x.
x= 3 or x= 4
Step 7: Substitute the values of xback into the expression we found for y
to find the corresponding values of y. For x= 3:y= 7 −3 = 4
For x= 4:y= 7 −4 = 3
Step 8: The solutions to the system of equations are (3,4) and (4,3).
Question 5
Question
Find the standard form of the equation of the ellipse with foci at (−2,0) and
(2,0) passing through the point (4,3).
Solution
Step 1: Find the center of the ellipse. To find the center of the ellipse, we
can take the average of the foci coordinates. The center of the ellipse is at the
midpoint of the foci: (−2+2
2,0+0
2)= (0,0).
Step 2: Find the distance between the foci. The distance between the foci
is 2(2) = 4 (since the foci have coordinates (±2,0)).
4
Step 3: Find the distance between one of the foci and the point (4,3). Using
the distance formula √(x2−x1)2+ (y2−y1)2, we have:
√(2 −4)2+ (0 −3)2=√(−2)2+ (−3)2=√4 + 9 = √13.
Step 4: Find the lengths of the semi-major axis (a) and the semi-minor axis
(b). The sum of the distances from the foci to any point on the ellipse is 2a
(the major axis length). Therefore, 2a= 4 =⇒a= 2. Also, a2=b2+c2
where cis the distance from the center to a focus. Since c= 2, we have
22=b2+ 22=⇒4 = b2+ 4 =⇒b2= 0 =⇒b= 0.
Step 5: Write the standard form of the equation of the ellipse. With the
center at the origin (0,0) and the values for aand b, the equation of the ellipse
is:
x2
4+y2
0= 1 .
Question 6
Question
Solve the equation 3x2+ 4xy +y2= 10 for xin terms of y.
Solution
Step 1: Rewrite the equation in standard form by completing the square for x.
3x2+ 4xy +y2= 10
3x2+ 4xy + 4 (y
2)2= 10 + 4 (y
2)2
3x2+ 4xy + 4 (y
2)2= 10 + 2y2
Step 2: Factor the trinomial.
(3x+ 2y)(x+ 2y) = 10 + 2y2
5
Step 3: Solve for xin terms of y.
(3x+ 2y)(x+ 2y) = 10 + 2y2
3x+ 2y=10 + 2y2
x+ 2y
3x+ 2y=10 + 2y2
x+ 2y
3x(x+ 2y)+2y(x+ 2y) = 10 + 2y2
3x2+ 6xy + 2xy + 4y2= 10 + 2y2
3x2+ 8xy + 4y2= 10 + 2y2
3x2+ 8xy + 4y2−2y2= 10
3x2+ 8xy + 2y2= 10
Step 4: Factor the trinomial.
(3x+y)(x+ 2y) = 10
Therefore, the solution to the equation 3x2+ 4xy +y2= 10 for xin terms
of yis x=10
3y+y.
Question 7
Question
Solve the system of equations:
{x2+y2= 25
x−y= 1
Solution
Step 1: Solve the second equation for x:
x=y+ 1
Step 2: Substitute the expression y+ 1 for xin the first equation:
(y+ 1)2+y2= 25
Step 3: Expand and simplify the equation:
y2+ 2y+1+y2= 25
6
2y2+ 2y−24 = 0
Step 4: Divide the equation by 2 to simplify:
y2+y−12 = 0
Step 5: Factor the quadratic equation:
(y+ 4)(y−3) = 0
Step 6: Set each factor to zero and solve for y:
y+ 4 = 0 =⇒y=−4
y−3 = 0 =⇒y= 3
Step 7: Use the values of yto find the corresponding values of x: For y=−4:
x=−4 + 1 =⇒x=−3
For y= 3:
x= 3 + 1 =⇒x= 4
Step 8: The solutions to the system of equations are:
(x, y) = (−3,−4),(4,3)
Question 8
Question
Solve the following system of equations:
{x2+y2= 25
x−y= 3
Solution
Step 1: Subtract the second equation from the first to eliminate y:
x2+y2−(x−y) = 25 −3
Step 2: Simplify the equation:
x2+y2−x+y= 22
Step 3: Simplify further by rearranging the terms:
x2−x+y2+y= 22
7
Step 4: Complete the square for both xand y: For x: Add (1/2)2= 1/4
inside the bracket:
x2−x+1
4+y2+y= 22 + 1
4
Step 5: Similarly, complete the square for y: Add (1/2)2= 1/4inside the
bracket:
x2−x+1
4+y2+y+1
4= 22 + 1
4+1
4
Step 6: Factor the left side:
(x−1
2)2
+(y+1
2)2
= 22 + 1
2
Step 7: Simplify the right side:
(x−1
2)2
+(y+1
2)2
=45
2
Step 8: Since the equation represents a circle, its center is at (1
2,−1
2)and
its radius is √45/2.
Thus, the solution to the system of equations is the point of intersection
between the circle x2+y2= 25 and the line x−y= 3.
Question 9
Question
Solve the following system of equations:
{x2+y2= 9
y=x2−2x
Solution
Step 1: Substitute the expression for yfrom the second equation into the first
equation.
x2+ (x2−2x)2= 9
Step 2: Simplify the equation.
x2+ (x4−4x3+ 4x2) = 9
Step 3: Combine like terms.
x4−4x3+ 5x2−9 = 0
Step 4: Factor the equation, if possible. In this case, factoring is not straight-
forward, so we’ll need to use numerical methods to find the solutions.
8
Step 5: Using a graphing calculator or numerical methods, we find that the
solutions to the equation are x=−1,x= 1,x≈1.78, and x≈2.78.
Step 6: Substitute these values back into the equation y=x2−2xto find
the corresponding yvalues.
Therefore, the solutions to the system of equations are (−1,3),(1,−1),
(1.78,−0.704), and (2.78,1.184).
Question 10
Question
Solve the following system of equations in terms of aand b:
{2x2−3y2=a
6x−5y=b
Solution
To solve this system of equations, we will use the substitution method to elim-
inate one of the variables.
Step 1: Solve the second equation for xin terms of y:We rearrange
the second equation to solve for x:
6x−5y=b
6x= 5y+b
x=5y+b
6
Step 2: Substitute xinto the first equation: Substitute the expression
for xinto the first equation of the system:
2(5y+b
6)2
−3y2=a
This simplifies to:
50y2+ 20by +b2
18 −3y2=a
Step 3: Simplify the equation and solve for y:Combine and simplify
the terms: 50y2+ 20by +b2
18 −54y2
18 =a
50y2+ 20by +b2−54y2= 18a
50y2−54y2+ 20by = 18a−b2
−4y2+ 20by = 18a−b2
9
−4y(y−5b) = 18a−b2
y(y−5b) = b2−18a
4
y2−5by =b2−18a
4
Step 4: Solve for xusing the obtained value for y:Substitute the
value of yback into the equation x=5y+b
6:
x=5(±√y2−5by) + b
6
x=5(±√b2
−18a
4) + b
6
Therefore, the solutions for xand yin terms of aand bare:
y=±√b2−18a
4, x =5(±√b2
−18a
4) + b
6
Question 11
Question
Solve the system of equations:
{4x2+y2= 16
x−2y= 3
Solution
To solve this system of equations, we will use substitution method.
Step 1: Solve the second equation for xin terms of y.
x= 3 + 2y
Step 2: Substitute xfrom the second equation into the first equation and
solve for y.
4(3 + 2y)2+y2= 16
4(9 + 12y+ 4y2) + y2= 16
36 + 48y+ 16y2+y2= 16
17y2+ 48y+ 20 = 0
10
(y+ 4)(17y+ 5) = 0
Therefore, y=−4or y=−5
17 .
Step 3: Substitute the values of yback into x= 3 + 2yto find the corre-
sponding values of x.
For y=−4, we get x= 3 + 2(−4) = 3 −8 = −5.
For y=−5
17 , we get x= 3 + 2 (−5
17 )= 3 −10
17 =51
17 −10
17 =41
17 .
Step 4: The solutions to the system of equations are x=−5, y =−4and
x=41
17 , y =−5
17 .
Question 12
Question
Find the standard form of the equation of a hyperbola given its foci at (−3,2)
and (3,2) and passing through the point (5,6).
Solution
Step 1: Find the center of the hyperbola.
The center of the hyperbola is the midpoint of the segment connecting the
foci. Use the midpoint formula:
(h, k) = (−3+3
2,2+2
2)
(h, k) = (0,2)
Step 2: Find the distance between the foci.
The distance between the foci of a hyperbola is 2a, where ais the distance
from the center to a vertex. Thus, we have:
√(3 −(−3))2+ (2 −2)2= 2a
6 = 2a
a= 3
Step 3: Find the equation of the hyperbola.
The standard form of the equation of a hyperbola centered at (h, k)with
vertices along the x-axis is:
(x−h)2
a2−(y−k)2
b2= 1
Since the hyperbola passes through the point (5,6), substitute x= 5 and y= 6
into the equation:
(5 −0)2
32−(6 −2)2
b2= 1
11
25/9−16/b2= 1
16/b2= 16/9
b2= 9
b= 3
Therefore, the standard form of the equation of the hyperbola is:
x2
9−(y−2)2
9= 1
Question 13
Question
Solve the following system of equations:
{x2+y2= 25
x+y= 7
Solution
Step 1: We can start by solving the second equation for xin terms of y:
x= 7 −y
Step 2: Substituting this expression for xinto the first equation gives:
(7 −y)2+y2= 25
Step 3: Simplifying the left side of the equation:
(49 −14y+y2) + y2= 25
49 −14y+ 2y2= 25
Step 4: Rearranging the equation:
2y2−14y+ 24 = 0
Step 5: Dividing the equation by 2 to simplify:
y2−7y+ 12 = 0
Step 6: Factoring the quadratic equation:
(y−4)(y−3) = 0
Step 7: Setting each factor to zero and solving for y:
y−4 = 0 =⇒y= 4
12
y−3 = 0 =⇒y= 3
Step 8: Now that we have the possible values for y, we can substitute them
back into the equation x= 7 −yto find the corresponding values of x: - For
y= 4:x= 7 −4 = 3 - For y= 3:x= 7 −3 = 4
Therefore, the solutions to the system of equations are (x, y) = (3,4) and
(x, y) = (4,3).
Question 14
Question
Solve the equation 3x2+ 4xy + 2y2−4x+ 4y+ 3 = 0 by classifying the conic
section it represents.
Solution
Step 1: To classify the conic section, we can rewrite the equation in matrix form
and find the eigenvalues of the resulting matrix. The matrix form of the given
equation is:
3 2 −2
2 2 2
−2 2 3
x
y
1
= 0
Step 2: To find the eigenvalues of the matrix, we set up and solve the
characteristic equation:
det
3−λ2−2
2 2 −λ2
−2 2 3 −λ
= 0
Step 3: Solving the characteristic equation, we find the eigenvalues λ1= 2,
λ2= 5, and λ3= 1.
Step 4: Since we have both positive and negative eigenvalues, the conic
section represented by the given equation is a hyperbola.
Question 15
Question
Solve the following system of equations:
{2x2+y2= 18
x2−2y2= 2
13
Solution
Step 1: Start by isolating y2in the first equation.
{y2= 18 −2x2
x2−2y2= 2
Step 2: Substitute y2= 18 −2x2into the second equation.
x2−2(18 −2x2) = 2
Step 3: Simplify and solve for x.
x2−36 + 4x2= 2
5x2= 38
x2=38
5
x=±√38
5
Step 4: Substitute xback into y2= 18 −2x2to find y.
y2= 18 −2(38
5)
y2= 18 −76
5
y2=90
5−76
5
y2=14
5
y=±√14
5
Step 5: The solutions to the system of equations are:
(√38
5,√14
5)and (−√38
5,−√14
5)
Question 16
Question
Solve the following system of equations:
{x2+ 2y2= 2
4x2−y2= 5
14
Solution
Step 1: First, we can rearrange the second equation to solve for y2:
4x2−y2= 5 =⇒y2= 4x2−5
Step 2: Substitute the expression for y2into the first equation:
x2+ 2(4x2−5) = 2
Step 3: Simplify the equation:
x2+ 8x2−10 = 2 =⇒9x2= 12 =⇒x2=4
3
Step 4: Take the square root of both sides to solve for x:
x=±√4
3=±2
√3=±2√3
3
Step 5: Substitute the values of xback into one of the original equations.
We’ll use the first equation:
{(2√3
3)2+ 2y2= 2
(7√3)2−y2= 5
Step 6: Simplify the equations:
{4
3+ 2y2= 2
21 −y2= 5
Step 7: Solve for yin the second equation:
y2= 21 −5 = 16 =⇒y=±4
Step 8: Solve for yusing y=±4in the first equation:
{4
3+ 2(4)2= 2 or 4
3+ 2(−4)2= 2
21 −(4)2= 5 or 21 −(−4)2= 5
Step 9: Simplify and solve for y:
{4
3+ 32 = 2 or 4
3+ 32 = 2
21 −16 = 5 or 21 −16 = 5
Step 10: Check our solutions: Both sets of solution should satisfy the original
system of equations:
{(2√3
3)2+ 2(4)2= 2
(7√3)2−42= 5 and {(−2√3
3)2+ 2(−4)2= 2
(−7√3)2−42= 5
Hence, the solutions are x=±2√3
3and y=±4.
15
Question 17
Question
Solve the following system of equations:
{x2+y2= 25
x−y= 1
Solution
Step 1: Solve the second equation for either variable. Let’s solve for xin terms
of y:
x=y+ 1
Step 2: Substitute xin terms of yinto the first equation:
(y+ 1)2+y2= 25
Step 3: Expand and simplify the equation:
y2+ 2y+1+y2= 25
2y2+ 2y−24 = 0
Step 4: Divide by 2 to simplify the equation:
y2+y−12 = 0
Step 5: Factor the quadratic equation:
(y+ 4)(y−3) = 0
Step 6: Set each factor to zero and solve for y:
y+ 4 = 0 =⇒y=−4
y−3 = 0 =⇒y= 3
Step 7: Substitute the values of yback into the equation x=y+ 1 to find
the corresponding values of x:
For y=−4 : x=−4 + 1 = −3
For y= 3 : x= 3 + 1 = 4
Step 8: Therefore, the solutions to the system of equations are:
{x=−3, y =−4
x= 4, y = 3
16
Question 18
Question
Solve the following system of equations:
{y=x2+ 4x
y= 3x+ 6
Solution
Step 1: We start by setting the two expressions for yequal to each other to find
the value(s) of x:
x2+ 4x= 3x+ 6
Step 2: Rearrange the equation and set it equal to zero by subtracting 3x
and 6from both sides:
x2+ 4x−3x−6 = 0
Step 3: Simplify the equation:
x2+x−6 = 0
Step 4: Factor the quadratic equation:
(x−2)(x+ 3) = 0
Step 5: Set each factor to zero and solve for x:
x−2 = 0 or x+ 3 = 0
Step 6: Solve for x:
x= 2 or x=−3
Step 7: Substitute the value of xback into the equation y=x2+ 4xto find
the corresponding values of y: For x= 2:
y= 22+ 4(2) = 4 + 8 = 12
So, when x= 2,y= 12.
For x=−3:
y= (−3)2+ 4(−3) = 9 −12 = −3
So, when x=−3,y=−3.
Step 8: The solution to the system of equations is therefore:
(x, y) = (2,12),(−3,−3)
17
Question 19
Question
Consider the equation of the hyperbola (x−1)2
9−(y+2)2
4= 1. Determine the
vertices, foci, and equations of the asymptotes of this hyperbola.
Solution
Step 1: First, identify key information from the given equation. The general
equation of a hyperbola centered at (h, k)is of the form (x−h)2
a2−(y−k)2
b2= 1
for a hyperbola opening horizontally or (y−k)2
a2−(x−h)2
b2= 1 for a hyperbola
opening vertically. In our case, the center of the hyperbola is at (1,−2),a2= 9,
and b2= 4.
Step 2: The vertices of the hyperbola are given by the points (h±a, k)for a
horizontally-opening hyperbola and (h, k±a)for a vertically-opening hyperbola.
So, the vertices are at (1 + 3,−2) = (4,−2) and (1 −3,−2) = (−2,−2).
Step 3: The distance from the center to the foci is given by c=√a2+b2.
In this case, c=√9 + 4 = √13. The foci are located at a distance of √13 units
from the center along the transverse axis, in this case, the x-axis. Therefore,
the foci are at (1 + √13,−2) and (1 −√13,−2).
Step 4: The slopes of the asymptotes of a hyperbola are given by ±b
a. In
this case, the slopes are ±2
3. Since the center of the hyperbola is at (1,−2), the
asymptotes will have the equation y+ 2 = ±2
3(x−1). Simplifying, we get the
equations of the asymptotes as y=2
3x−8
3and y=−2
3x−4
3.
Question 20
Question
Consider the conic section defined by the equation x2
4+y2
9−2x+ 6y+ 1 = 0.
Find the standard form of this conic section, determine its type, and sketch its
graph.
Solution
Step 1: Rearrange the equation into standard form by completing the square
for xand y:
x2
4+y2
9−2x+ 6y+ 1 = 0
x2
4−2x+y2
9+ 6y=−1
x2
4−2x+1+y2
9+ 6y+ 9 = −1+1+9
18
(x
2−1)2+(y
3+ 2)2= 9
Step 2: The standard form of the conic section is (x
2−1)2+(y
3+ 2)2= 32.
Step 3: Since the coefficients of x2and y2are positive and equal, the conic
section is a circle.
Step 4: To sketch the graph, we note that the center of the circle is at (2,−2)
and the radius is 3. Therefore, plot the point (2,−2) as the center and draw a
circle of radius 3.
Step 5: Sketch the circle on the coordinate plane.
Thus, the standard form of the conic section is (x
2−1)2+(y
3+ 2)2= 9, it
is a circle, and the graph is centered at (2, -2) with a radius of 3.
Question 21
Question
Solve the system of equations:
{x2+ 4y2= 16
x2−2x+ 2y2−4y= 7
Solution
Step 1: Begin by rewriting the second equation in the system by completing the
square for both the xand yterms:
x2+ 4y2= 16
(x2−2x) + (2y2−4y) = 7
(x2−2x+ 1) + (2(y2−2y+ 1)) = 7 + 1 + 2
(x−1)2+ 2(y−1)2= 10
Step 2: Make a substitution by letting X=x−1and Y=y−1so that the
equation becomes:
{x2+ 4y2= 16
X2+ 2Y2= 10
Step 3: Multiply the second equation by 4:
{x2+ 4y2= 16
4X2+ 8Y2= 40
Step 4: Add the first equation to the multiple of the second equation:
5X2= 56 =⇒X2=56
5
19
Step 5: Solve for X:
X=±√56
5=±2√14
5
Step 6: Substitute the value of Xback in and solve for Y:
2Y2= 10 −4( 14
5) = 10 −11.2 = −1.2 =⇒Y2=−0.6
Step 7: Since Y2is negative, there are no real solutions to this system of
equations.
Question 22
Question
Solve the system of equations:
{x2−y2= 16
xy =−8
Solution
Step 1: Let’s solve the second equation for yin terms of x.
xy =−8 =⇒y=−8
x
Step 2: Substitute y=−8
xinto the first equation.
x2−(−8
x)2
= 16
Step 3: Simplify the equation by squaring and combining terms.
x2−64
x2= 16
Step 4: Multiply through by x2to clear the fraction.
x4−64 = 16x2
Step 5: Rearrange the equation into standard form and solve for x.
x4−16x2−64 = 0
Step 6: Let u=x2to simplify the equation.
u2−16u−64 = 0
20
Step 7: Factor the quadratic equation.
(u−8)(u+ 8) = 0
Step 8: Set each factor to zero and solve for u.
u= 8 or u=−8
Step 9: Substitute back u=x2to find possible values for x. For u= 8:
x2= 8 =⇒x=±2√2. For u=−8: No real solutions.
Step 10: Finally, substitute the values of xback into y=−8
xto find the
corresponding yvalues. Thus, the solutions to the system of equations are:
(x, y) = (2√2,−4√2) and (−2√2,4√2) .
Question 23
Question
Solve the following system of equations:
{4x2+ 9y2= 36
x2−y2= 1
Solution
Step 1: We can start by solving the second equation for y2:
x2−y2= 1
y2=x2−1
Step 2: Now we can substitute y2=x2−1into the first equation:
4x2+ 9(x2−1) = 36
Step 3: Simplify the equation:
4x2+ 9x2−9 = 36
13x2−9 = 36
Step 4: Add 9 to both sides:
13x2= 45
Step 5: Divide by 13:
x2=45
13
21
Step 6: Taking the square root of both sides, we get two possible solutions
for x:
x=±√45
13
Step 7: Substituting back into y2=x2−1to find y: For x=√45
13 :
y2=45
13 −1 = 32
13
y=±√32
13
For x=−√45
13 :
y2=45
13 −1 = 32
13
y=±√32
13
Therefore, the solutions to the system of equations are:
(x, y) = (√45
13,±√32
13)and (−√45
13,±√32
13)
Question 24
Question
Let A(1,3),B(5,6), and C(2,−2) be three points in the plane. Determine the
equation of the circle passing through these three points.
Solution
Step 1: Find the equations of the perpendicular bisectors of the line segments
AB and BC.
To find the equation of the perpendicular bisector of the line segment AB,
we first find the midpoint Mof AB, which is given by:
M(1+5
2,3+6
2)=M(3,4.5).
The slope of the line passing through Aand Bis:
mAB =6−3
5−1=3
4.
22
Therefore, the slope of the perpendicular bisector of AB is −4
3. Since this line
passes through the midpoint M(3,4.5), its equation can be written as:
y−4.5 = −4
3(x−3).
Similarly, for the line passing through Band C, first find the midpoint N
of BC, which is given by:
N(5+2
2,6+(−2)
2)=N(7
2,2).
The slope of the line passing through Band Cis:
mBC =2−6
2−5=−4
3.
Thus, the slope of the perpendicular bisector of BC is 3
4. Hence, its equation
is:
y−2 = 3
4(x−7
2).
Step 2: Solve the system of equations formed by the perpendicular bisectors
calculated in Step 1.
Solve the system of equations
{y−4.5 = −4
3(x−3)
y−2 = 3
4(x−7
2)
to find the point Oat which the perpendicular bisectors intersect.
Step 3: Determine the radius of the circle.
Calculate the distance between the point Oand any of the given points A,
B, or Cto find the radius of the circle.
Step 4: Write the equation of the circle.
Finally, using the center Oand the radius, write the equation of the circle
passing through points A,B, and C.
Question 25
Question
Find the standard form of the equation of the hyperbola with vertices at (−5,0)
and (5,0), and foci at (−7,0) and (7,0).
Solution
Step 1: Find the center of the hyperbola. The center of the hyperbola is the
midpoint of the segment connecting the vertices. The midpoint formula is given
by (x1+x2
2,y1+y2
2). Therefore, the center is ((5 + (−5))/2,(0 + 0)/2) = (0,0).
23
Step 2: Find the distance from the center to a vertex. This distance is called
the distance from the center to the vertex, which is a= 5 −0 = 5.
Step 3: Find the distance from the center to a focus. This distance is called
the distance from the center to the focus, which is c= 7 −0 = 7.
Step 4: Find the value of busing the relationship c2=a2+b2. Substitute
a= 5 and c= 7 into the equation and solve for b:
72= 52+b2
49 = 25 + b2
b2= 24
b=√24 = 2√6
Step 5: Write the equation of the hyperbola using the given information.
The standard form of the equation of a hyperbola centered at the origin is
x2
a2−y2
b2= 1. Substituting the values a= 5 and b= 2√6into the equation, we
have: x2
52−y2
(2√6)2= 1
x2
25 −y2
24 = 1
Therefore, the standard form of the equation of the hyperbola is x2
25 −y2
24 = 1.
24
Question 2
Question
Find the standard form of the equation of the parabola with focus F(0,3) and
directrix y=−3.
Solution
Step 1: Recall the definition of a parabola. A parabola is the set of all points
that are equidistant from a fixed point called the focus and a fixed line called
the directrix.
Step 2: Given that the focus is F(0,3) and the directrix is y=−3, we can
see that the vertex is the midpoint between the focus and the directrix. So, the
vertex is at (0,0).
Step 3: Since the directrix is a horizontal line, we have a vertical parabola
opening either upwards or downwards. Since the focus is above the directrix,
the parabola opens upwards.
Step 4: The distance between the vertex and the focus is the same as the
distance between the vertex and the directrix. This distance is called the focal
length. In this case, the focal length is 3units.
Step 5: The standard form of the equation of a parabola with vertex at (h, k)
and focal length |p|is given by:
(x−h)2= 4p(y−k)
Step 6: Plugging in the values h= 0,k= 0, and p= 3 into the standard
form equation, we get:
x2= 12y
Therefore, the standard form of the equation of the parabola is x2= 12y.
Question 3
Question
Solve the following system of equations:
{x2+y2= 25
x+ 2y= 10
Solution
Step 1: Solve the second equation for xin terms of y:
x= 10 −2y
2
Step 2: Substitute x= 10 −2yinto the first equation:
(10 −2y)2+y2= 25
Step 3: Expand and simplify the equation:
100 −40y+ 4y2+y2= 25
Step 4: Rearrange the equation into standard quadratic form:
5y2−40y+ 75 = 0
Step 5: Solve the quadratic equation for yusing the quadratic formula:
y=−(−40) ±√(−40)2−4(5)(75)
2(5)
Step 6: Simplify the expression under the square root:
y=40 ±√1600 −1500
10
y=40 ±√100
10
Step 7: Solve for y:
y=40 ±10
10
y= 5,3
Step 8: Substitute y= 5 into x= 10 −2y:
x= 10 −2(5) = 0
Step 9: Substitute y= 3 into x= 10 −2y:
x= 10 −2(3) = 4
Step 10: The solutions to the system are (0,5) and (4,3).
Question 4
Question
Solve the following system of equations:
{x2+y2= 25
x+y= 7
3
Solution
Step 1: Rewrite the second equation in terms of one variable. Solve for either
xor yin terms of the other variable. We will solve for yin terms of x.
y= 7 −x
Step 2: Substitute the expression for yinto the first equation.
x2+ (7 −x)2= 25
Step 3: Expand and simplify the equation.
x2+ 49 −14x+x2= 25
2x2−14x+ 24 = 0
Step 4: Rearrange the equation into standard quadratic form.
x2−7x+ 12 = 0
Step 5: Factor the quadratic equation.
(x−3)(x−4) = 0
Step 6: Set each factor to zero and solve for x.
x= 3 or x= 4
Step 7: Substitute the values of xback into the expression we found for y
to find the corresponding values of y. For x= 3:y= 7 −3 = 4
For x= 4:y= 7 −4 = 3
Step 8: The solutions to the system of equations are (3,4) and (4,3).
Question 5
Question
Find the standard form of the equation of the ellipse with foci at (−2,0) and
(2,0) passing through the point (4,3).
Solution
Step 1: Find the center of the ellipse. To find the center of the ellipse, we
can take the average of the foci coordinates. The center of the ellipse is at the
midpoint of the foci: (−2+2
2,0+0
2)= (0,0).
Step 2: Find the distance between the foci. The distance between the foci
is 2(2) = 4 (since the foci have coordinates (±2,0)).
4
Step 3: Find the distance between one of the foci and the point (4,3). Using
the distance formula √(x2−x1)2+ (y2−y1)2, we have:
√(2 −4)2+ (0 −3)2=√(−2)2+ (−3)2=√4 + 9 = √13.
Step 4: Find the lengths of the semi-major axis (a) and the semi-minor axis
(b). The sum of the distances from the foci to any point on the ellipse is 2a
(the major axis length). Therefore, 2a= 4 =⇒a= 2. Also, a2=b2+c2
where cis the distance from the center to a focus. Since c= 2, we have
22=b2+ 22=⇒4 = b2+ 4 =⇒b2= 0 =⇒b= 0.
Step 5: Write the standard form of the equation of the ellipse. With the
center at the origin (0,0) and the values for aand b, the equation of the ellipse
is:
x2
4+y2
0= 1 .
Question 6
Question
Solve the equation 3x2+ 4xy +y2= 10 for xin terms of y.
Solution
Step 1: Rewrite the equation in standard form by completing the square for x.
3x2+ 4xy +y2= 10
3x2+ 4xy + 4 (y
2)2= 10 + 4 (y
2)2
3x2+ 4xy + 4 (y
2)2= 10 + 2y2
Step 2: Factor the trinomial.
(3x+ 2y)(x+ 2y) = 10 + 2y2
5
Step 3: Solve for xin terms of y.
(3x+ 2y)(x+ 2y) = 10 + 2y2
3x+ 2y=10 + 2y2
x+ 2y
3x+ 2y=10 + 2y2
x+ 2y
3x(x+ 2y)+2y(x+ 2y) = 10 + 2y2
3x2+ 6xy + 2xy + 4y2= 10 + 2y2
3x2+ 8xy + 4y2= 10 + 2y2
3x2+ 8xy + 4y2−2y2= 10
3x2+ 8xy + 2y2= 10
Step 4: Factor the trinomial.
(3x+y)(x+ 2y) = 10
Therefore, the solution to the equation 3x2+ 4xy +y2= 10 for xin terms
of yis x=10
3y+y.
Question 7
Question
Solve the system of equations:
{x2+y2= 25
x−y= 1
Solution
Step 1: Solve the second equation for x:
x=y+ 1
Step 2: Substitute the expression y+ 1 for xin the first equation:
(y+ 1)2+y2= 25
Step 3: Expand and simplify the equation:
y2+ 2y+1+y2= 25
6
2y2+ 2y−24 = 0
Step 4: Divide the equation by 2 to simplify:
y2+y−12 = 0
Step 5: Factor the quadratic equation:
(y+ 4)(y−3) = 0
Step 6: Set each factor to zero and solve for y:
y+ 4 = 0 =⇒y=−4
y−3 = 0 =⇒y= 3
Step 7: Use the values of yto find the corresponding values of x: For y=−4:
x=−4 + 1 =⇒x=−3
For y= 3:
x= 3 + 1 =⇒x= 4
Step 8: The solutions to the system of equations are:
(x, y) = (−3,−4),(4,3)
Question 8
Question
Solve the following system of equations:
{x2+y2= 25
x−y= 3
Solution
Step 1: Subtract the second equation from the first to eliminate y:
x2+y2−(x−y) = 25 −3
Step 2: Simplify the equation:
x2+y2−x+y= 22
Step 3: Simplify further by rearranging the terms:
x2−x+y2+y= 22
7
Step 4: Complete the square for both xand y: For x: Add (1/2)2= 1/4
inside the bracket:
x2−x+1
4+y2+y= 22 + 1
4
Step 5: Similarly, complete the square for y: Add (1/2)2= 1/4inside the
bracket:
x2−x+1
4+y2+y+1
4= 22 + 1
4+1
4
Step 6: Factor the left side:
(x−1
2)2
+(y+1
2)2
= 22 + 1
2
Step 7: Simplify the right side:
(x−1
2)2
+(y+1
2)2
=45
2
Step 8: Since the equation represents a circle, its center is at (1
2,−1
2)and
its radius is √45/2.
Thus, the solution to the system of equations is the point of intersection
between the circle x2+y2= 25 and the line x−y= 3.
Question 9
Question
Solve the following system of equations:
{x2+y2= 9
y=x2−2x
Solution
Step 1: Substitute the expression for yfrom the second equation into the first
equation.
x2+ (x2−2x)2= 9
Step 2: Simplify the equation.
x2+ (x4−4x3+ 4x2) = 9
Step 3: Combine like terms.
x4−4x3+ 5x2−9 = 0
Step 4: Factor the equation, if possible. In this case, factoring is not straight-
forward, so we’ll need to use numerical methods to find the solutions.
8
Step 5: Using a graphing calculator or numerical methods, we find that the
solutions to the equation are x=−1,x= 1,x≈1.78, and x≈2.78.
Step 6: Substitute these values back into the equation y=x2−2xto find
the corresponding yvalues.
Therefore, the solutions to the system of equations are (−1,3),(1,−1),
(1.78,−0.704), and (2.78,1.184).
Question 10
Question
Solve the following system of equations in terms of aand b:
{2x2−3y2=a
6x−5y=b
Solution
To solve this system of equations, we will use the substitution method to elim-
inate one of the variables.
Step 1: Solve the second equation for xin terms of y:We rearrange
the second equation to solve for x:
6x−5y=b
6x= 5y+b
x=5y+b
6
Step 2: Substitute xinto the first equation: Substitute the expression
for xinto the first equation of the system:
2(5y+b
6)2
−3y2=a
This simplifies to:
50y2+ 20by +b2
18 −3y2=a
Step 3: Simplify the equation and solve for y:Combine and simplify
the terms: 50y2+ 20by +b2
18 −54y2
18 =a
50y2+ 20by +b2−54y2= 18a
50y2−54y2+ 20by = 18a−b2
−4y2+ 20by = 18a−b2
9
−4y(y−5b) = 18a−b2
y(y−5b) = b2−18a
4
y2−5by =b2−18a
4
Step 4: Solve for xusing the obtained value for y:Substitute the
value of yback into the equation x=5y+b
6:
x=5(±√y2−5by) + b
6
x=5(±√b2
−18a
4) + b
6
Therefore, the solutions for xand yin terms of aand bare:
y=±√b2−18a
4, x =5(±√b2
−18a
4) + b
6
Question 11
Question
Solve the system of equations:
{4x2+y2= 16
x−2y= 3
Solution
To solve this system of equations, we will use substitution method.
Step 1: Solve the second equation for xin terms of y.
x= 3 + 2y
Step 2: Substitute xfrom the second equation into the first equation and
solve for y.
4(3 + 2y)2+y2= 16
4(9 + 12y+ 4y2) + y2= 16
36 + 48y+ 16y2+y2= 16
17y2+ 48y+ 20 = 0
10
(y+ 4)(17y+ 5) = 0
Therefore, y=−4or y=−5
17 .
Step 3: Substitute the values of yback into x= 3 + 2yto find the corre-
sponding values of x.
For y=−4, we get x= 3 + 2(−4) = 3 −8 = −5.
For y=−5
17 , we get x= 3 + 2 (−5
17 )= 3 −10
17 =51
17 −10
17 =41
17 .
Step 4: The solutions to the system of equations are x=−5, y =−4and
x=41
17 , y =−5
17 .
Question 12
Question
Find the standard form of the equation of a hyperbola given its foci at (−3,2)
and (3,2) and passing through the point (5,6).
Solution
Step 1: Find the center of the hyperbola.
The center of the hyperbola is the midpoint of the segment connecting the
foci. Use the midpoint formula:
(h, k) = (−3+3
2,2+2
2)
(h, k) = (0,2)
Step 2: Find the distance between the foci.
The distance between the foci of a hyperbola is 2a, where ais the distance
from the center to a vertex. Thus, we have:
√(3 −(−3))2+ (2 −2)2= 2a
6 = 2a
a= 3
Step 3: Find the equation of the hyperbola.
The standard form of the equation of a hyperbola centered at (h, k)with
vertices along the x-axis is:
(x−h)2
a2−(y−k)2
b2= 1
Since the hyperbola passes through the point (5,6), substitute x= 5 and y= 6
into the equation:
(5 −0)2
32−(6 −2)2
b2= 1
11
25/9−16/b2= 1
16/b2= 16/9
b2= 9
b= 3
Therefore, the standard form of the equation of the hyperbola is:
x2
9−(y−2)2
9= 1
Question 13
Question
Solve the following system of equations:
{x2+y2= 25
x+y= 7
Solution
Step 1: We can start by solving the second equation for xin terms of y:
x= 7 −y
Step 2: Substituting this expression for xinto the first equation gives:
(7 −y)2+y2= 25
Step 3: Simplifying the left side of the equation:
(49 −14y+y2) + y2= 25
49 −14y+ 2y2= 25
Step 4: Rearranging the equation:
2y2−14y+ 24 = 0
Step 5: Dividing the equation by 2 to simplify:
y2−7y+ 12 = 0
Step 6: Factoring the quadratic equation:
(y−4)(y−3) = 0
Step 7: Setting each factor to zero and solving for y:
y−4 = 0 =⇒y= 4
12
y−3 = 0 =⇒y= 3
Step 8: Now that we have the possible values for y, we can substitute them
back into the equation x= 7 −yto find the corresponding values of x: - For
y= 4:x= 7 −4 = 3 - For y= 3:x= 7 −3 = 4
Therefore, the solutions to the system of equations are (x, y) = (3,4) and
(x, y) = (4,3).
Question 14
Question
Solve the equation 3x2+ 4xy + 2y2−4x+ 4y+ 3 = 0 by classifying the conic
section it represents.
Solution
Step 1: To classify the conic section, we can rewrite the equation in matrix form
and find the eigenvalues of the resulting matrix. The matrix form of the given
equation is:
3 2 −2
2 2 2
−2 2 3
x
y
1
= 0
Step 2: To find the eigenvalues of the matrix, we set up and solve the
characteristic equation:
det
3−λ2−2
2 2 −λ2
−2 2 3 −λ
= 0
Step 3: Solving the characteristic equation, we find the eigenvalues λ1= 2,
λ2= 5, and λ3= 1.
Step 4: Since we have both positive and negative eigenvalues, the conic
section represented by the given equation is a hyperbola.
Question 15
Question
Solve the following system of equations:
{2x2+y2= 18
x2−2y2= 2
13
Solution
Step 1: Start by isolating y2in the first equation.
{y2= 18 −2x2
x2−2y2= 2
Step 2: Substitute y2= 18 −2x2into the second equation.
x2−2(18 −2x2) = 2
Step 3: Simplify and solve for x.
x2−36 + 4x2= 2
5x2= 38
x2=38
5
x=±√38
5
Step 4: Substitute xback into y2= 18 −2x2to find y.
y2= 18 −2(38
5)
y2= 18 −76
5
y2=90
5−76
5
y2=14
5
y=±√14
5
Step 5: The solutions to the system of equations are:
(√38
5,√14
5)and (−√38
5,−√14
5)
Question 16
Question
Solve the following system of equations:
{x2+ 2y2= 2
4x2−y2= 5
14
Solution
Step 1: First, we can rearrange the second equation to solve for y2:
4x2−y2= 5 =⇒y2= 4x2−5
Step 2: Substitute the expression for y2into the first equation:
x2+ 2(4x2−5) = 2
Step 3: Simplify the equation:
x2+ 8x2−10 = 2 =⇒9x2= 12 =⇒x2=4
3
Step 4: Take the square root of both sides to solve for x:
x=±√4
3=±2
√3=±2√3
3
Step 5: Substitute the values of xback into one of the original equations.
We’ll use the first equation:
{(2√3
3)2+ 2y2= 2
(7√3)2−y2= 5
Step 6: Simplify the equations:
{4
3+ 2y2= 2
21 −y2= 5
Step 7: Solve for yin the second equation:
y2= 21 −5 = 16 =⇒y=±4
Step 8: Solve for yusing y=±4in the first equation:
{4
3+ 2(4)2= 2 or 4
3+ 2(−4)2= 2
21 −(4)2= 5 or 21 −(−4)2= 5
Step 9: Simplify and solve for y:
{4
3+ 32 = 2 or 4
3+ 32 = 2
21 −16 = 5 or 21 −16 = 5
Step 10: Check our solutions: Both sets of solution should satisfy the original
system of equations:
{(2√3
3)2+ 2(4)2= 2
(7√3)2−42= 5 and {(−2√3
3)2+ 2(−4)2= 2
(−7√3)2−42= 5
Hence, the solutions are x=±2√3
3and y=±4.
15
Question 17
Question
Solve the following system of equations:
{x2+y2= 25
x−y= 1
Solution
Step 1: Solve the second equation for either variable. Let’s solve for xin terms
of y:
x=y+ 1
Step 2: Substitute xin terms of yinto the first equation:
(y+ 1)2+y2= 25
Step 3: Expand and simplify the equation:
y2+ 2y+1+y2= 25
2y2+ 2y−24 = 0
Step 4: Divide by 2 to simplify the equation:
y2+y−12 = 0
Step 5: Factor the quadratic equation:
(y+ 4)(y−3) = 0
Step 6: Set each factor to zero and solve for y:
y+ 4 = 0 =⇒y=−4
y−3 = 0 =⇒y= 3
Step 7: Substitute the values of yback into the equation x=y+ 1 to find
the corresponding values of x:
For y=−4 : x=−4 + 1 = −3
For y= 3 : x= 3 + 1 = 4
Step 8: Therefore, the solutions to the system of equations are:
{x=−3, y =−4
x= 4, y = 3
16
Question 18
Question
Solve the following system of equations:
{y=x2+ 4x
y= 3x+ 6
Solution
Step 1: We start by setting the two expressions for yequal to each other to find
the value(s) of x:
x2+ 4x= 3x+ 6
Step 2: Rearrange the equation and set it equal to zero by subtracting 3x
and 6from both sides:
x2+ 4x−3x−6 = 0
Step 3: Simplify the equation:
x2+x−6 = 0
Step 4: Factor the quadratic equation:
(x−2)(x+ 3) = 0
Step 5: Set each factor to zero and solve for x:
x−2 = 0 or x+ 3 = 0
Step 6: Solve for x:
x= 2 or x=−3
Step 7: Substitute the value of xback into the equation y=x2+ 4xto find
the corresponding values of y: For x= 2:
y= 22+ 4(2) = 4 + 8 = 12
So, when x= 2,y= 12.
For x=−3:
y= (−3)2+ 4(−3) = 9 −12 = −3
So, when x=−3,y=−3.
Step 8: The solution to the system of equations is therefore:
(x, y) = (2,12),(−3,−3)
17
Question 19
Question
Consider the equation of the hyperbola (x−1)2
9−(y+2)2
4= 1. Determine the
vertices, foci, and equations of the asymptotes of this hyperbola.
Solution
Step 1: First, identify key information from the given equation. The general
equation of a hyperbola centered at (h, k)is of the form (x−h)2
a2−(y−k)2
b2= 1
for a hyperbola opening horizontally or (y−k)2
a2−(x−h)2
b2= 1 for a hyperbola
opening vertically. In our case, the center of the hyperbola is at (1,−2),a2= 9,
and b2= 4.
Step 2: The vertices of the hyperbola are given by the points (h±a, k)for a
horizontally-opening hyperbola and (h, k±a)for a vertically-opening hyperbola.
So, the vertices are at (1 + 3,−2) = (4,−2) and (1 −3,−2) = (−2,−2).
Step 3: The distance from the center to the foci is given by c=√a2+b2.
In this case, c=√9 + 4 = √13. The foci are located at a distance of √13 units
from the center along the transverse axis, in this case, the x-axis. Therefore,
the foci are at (1 + √13,−2) and (1 −√13,−2).
Step 4: The slopes of the asymptotes of a hyperbola are given by ±b
a. In
this case, the slopes are ±2
3. Since the center of the hyperbola is at (1,−2), the
asymptotes will have the equation y+ 2 = ±2
3(x−1). Simplifying, we get the
equations of the asymptotes as y=2
3x−8
3and y=−2
3x−4
3.
Question 20
Question
Consider the conic section defined by the equation x2
4+y2
9−2x+ 6y+ 1 = 0.
Find the standard form of this conic section, determine its type, and sketch its
graph.
Solution
Step 1: Rearrange the equation into standard form by completing the square
for xand y:
x2
4+y2
9−2x+ 6y+ 1 = 0
x2
4−2x+y2
9+ 6y=−1
x2
4−2x+1+y2
9+ 6y+ 9 = −1+1+9
18
(x
2−1)2+(y
3+ 2)2= 9
Step 2: The standard form of the conic section is (x
2−1)2+(y
3+ 2)2= 32.
Step 3: Since the coefficients of x2and y2are positive and equal, the conic
section is a circle.
Step 4: To sketch the graph, we note that the center of the circle is at (2,−2)
and the radius is 3. Therefore, plot the point (2,−2) as the center and draw a
circle of radius 3.
Step 5: Sketch the circle on the coordinate plane.
Thus, the standard form of the conic section is (x
2−1)2+(y
3+ 2)2= 9, it
is a circle, and the graph is centered at (2, -2) with a radius of 3.
Question 21
Question
Solve the system of equations:
{x2+ 4y2= 16
x2−2x+ 2y2−4y= 7
Solution
Step 1: Begin by rewriting the second equation in the system by completing the
square for both the xand yterms:
x2+ 4y2= 16
(x2−2x) + (2y2−4y) = 7
(x2−2x+ 1) + (2(y2−2y+ 1)) = 7 + 1 + 2
(x−1)2+ 2(y−1)2= 10
Step 2: Make a substitution by letting X=x−1and Y=y−1so that the
equation becomes:
{x2+ 4y2= 16
X2+ 2Y2= 10
Step 3: Multiply the second equation by 4:
{x2+ 4y2= 16
4X2+ 8Y2= 40
Step 4: Add the first equation to the multiple of the second equation:
5X2= 56 =⇒X2=56
5
19
Step 5: Solve for X:
X=±√56
5=±2√14
5
Step 6: Substitute the value of Xback in and solve for Y:
2Y2= 10 −4( 14
5) = 10 −11.2 = −1.2 =⇒Y2=−0.6
Step 7: Since Y2is negative, there are no real solutions to this system of
equations.
Question 22
Question
Solve the system of equations:
{x2−y2= 16
xy =−8
Solution
Step 1: Let’s solve the second equation for yin terms of x.
xy =−8 =⇒y=−8
x
Step 2: Substitute y=−8
xinto the first equation.
x2−(−8
x)2
= 16
Step 3: Simplify the equation by squaring and combining terms.
x2−64
x2= 16
Step 4: Multiply through by x2to clear the fraction.
x4−64 = 16x2
Step 5: Rearrange the equation into standard form and solve for x.
x4−16x2−64 = 0
Step 6: Let u=x2to simplify the equation.
u2−16u−64 = 0
20
Step 7: Factor the quadratic equation.
(u−8)(u+ 8) = 0
Step 8: Set each factor to zero and solve for u.
u= 8 or u=−8
Step 9: Substitute back u=x2to find possible values for x. For u= 8:
x2= 8 =⇒x=±2√2. For u=−8: No real solutions.
Step 10: Finally, substitute the values of xback into y=−8
xto find the
corresponding yvalues. Thus, the solutions to the system of equations are:
(x, y) = (2√2,−4√2) and (−2√2,4√2) .
Question 23
Question
Solve the following system of equations:
{4x2+ 9y2= 36
x2−y2= 1
Solution
Step 1: We can start by solving the second equation for y2:
x2−y2= 1
y2=x2−1
Step 2: Now we can substitute y2=x2−1into the first equation:
4x2+ 9(x2−1) = 36
Step 3: Simplify the equation:
4x2+ 9x2−9 = 36
13x2−9 = 36
Step 4: Add 9 to both sides:
13x2= 45
Step 5: Divide by 13:
x2=45
13
21
Step 6: Taking the square root of both sides, we get two possible solutions
for x:
x=±√45
13
Step 7: Substituting back into y2=x2−1to find y: For x=√45
13 :
y2=45
13 −1 = 32
13
y=±√32
13
For x=−√45
13 :
y2=45
13 −1 = 32
13
y=±√32
13
Therefore, the solutions to the system of equations are:
(x, y) = (√45
13,±√32
13)and (−√45
13,±√32
13)
Question 24
Question
Let A(1,3),B(5,6), and C(2,−2) be three points in the plane. Determine the
equation of the circle passing through these three points.
Solution
Step 1: Find the equations of the perpendicular bisectors of the line segments
AB and BC.
To find the equation of the perpendicular bisector of the line segment AB,
we first find the midpoint Mof AB, which is given by:
M(1+5
2,3+6
2)=M(3,4.5).
The slope of the line passing through Aand Bis:
mAB =6−3
5−1=3
4.
22
Therefore, the slope of the perpendicular bisector of AB is −4
3. Since this line
passes through the midpoint M(3,4.5), its equation can be written as:
y−4.5 = −4
3(x−3).
Similarly, for the line passing through Band C, first find the midpoint N
of BC, which is given by:
N(5+2
2,6+(−2)
2)=N(7
2,2).
The slope of the line passing through Band Cis:
mBC =2−6
2−5=−4
3.
Thus, the slope of the perpendicular bisector of BC is 3
4. Hence, its equation
is:
y−2 = 3
4(x−7
2).
Step 2: Solve the system of equations formed by the perpendicular bisectors
calculated in Step 1.
Solve the system of equations
{y−4.5 = −4
3(x−3)
y−2 = 3
4(x−7
2)
to find the point Oat which the perpendicular bisectors intersect.
Step 3: Determine the radius of the circle.
Calculate the distance between the point Oand any of the given points A,
B, or Cto find the radius of the circle.
Step 4: Write the equation of the circle.
Finally, using the center Oand the radius, write the equation of the circle
passing through points A,B, and C.
Question 25
Question
Find the standard form of the equation of the hyperbola with vertices at (−5,0)
and (5,0), and foci at (−7,0) and (7,0).
Solution
Step 1: Find the center of the hyperbola. The center of the hyperbola is the
midpoint of the segment connecting the vertices. The midpoint formula is given
by (x1+x2
2,y1+y2
2). Therefore, the center is ((5 + (−5))/2,(0 + 0)/2) = (0,0).
23
Step 2: Find the distance from the center to a vertex. This distance is called
the distance from the center to the vertex, which is a= 5 −0 = 5.
Step 3: Find the distance from the center to a focus. This distance is called
the distance from the center to the focus, which is c= 7 −0 = 7.
Step 4: Find the value of busing the relationship c2=a2+b2. Substitute
a= 5 and c= 7 into the equation and solve for b:
72= 52+b2
49 = 25 + b2
b2= 24
b=√24 = 2√6
Step 5: Write the equation of the hyperbola using the given information.
The standard form of the equation of a hyperbola centered at the origin is
x2
a2−y2
b2= 1. Substituting the values a= 5 and b= 2√6into the equation, we
have: x2
52−y2
(2√6)2= 1
x2
25 −y2
24 = 1
Therefore, the standard form of the equation of the hyperbola is x2
25 −y2
24 = 1.
24
Question 2
Question
Find the standard form of the equation of the parabola with focus F(0,3) and
directrix y=−3.
Solution
Step 1: Recall the definition of a parabola. A parabola is the set of all points
that are equidistant from a fixed point called the focus and a fixed line called
the directrix.
Step 2: Given that the focus is F(0,3) and the directrix is y=−3, we can
see that the vertex is the midpoint between the focus and the directrix. So, the
vertex is at (0,0).
Step 3: Since the directrix is a horizontal line, we have a vertical parabola
opening either upwards or downwards. Since the focus is above the directrix,
the parabola opens upwards.
Step 4: The distance between the vertex and the focus is the same as the
distance between the vertex and the directrix. This distance is called the focal
length. In this case, the focal length is 3units.
Step 5: The standard form of the equation of a parabola with vertex at (h, k)
and focal length |p|is given by:
(x−h)2= 4p(y−k)
Step 6: Plugging in the values h= 0,k= 0, and p= 3 into the standard
form equation, we get:
x2= 12y
Therefore, the standard form of the equation of the parabola is x2= 12y.
Question 3
Question
Solve the following system of equations:
{x2+y2= 25
x+ 2y= 10
Solution
Step 1: Solve the second equation for xin terms of y:
x= 10 −2y
2
Step 2: Substitute x= 10 −2yinto the first equation:
(10 −2y)2+y2= 25
Step 3: Expand and simplify the equation:
100 −40y+ 4y2+y2= 25
Step 4: Rearrange the equation into standard quadratic form:
5y2−40y+ 75 = 0
Step 5: Solve the quadratic equation for yusing the quadratic formula:
y=−(−40) ±√(−40)2−4(5)(75)
2(5)
Step 6: Simplify the expression under the square root:
y=40 ±√1600 −1500
10
y=40 ±√100
10
Step 7: Solve for y:
y=40 ±10
10
y= 5,3
Step 8: Substitute y= 5 into x= 10 −2y:
x= 10 −2(5) = 0
Step 9: Substitute y= 3 into x= 10 −2y:
x= 10 −2(3) = 4
Step 10: The solutions to the system are (0,5) and (4,3).
Question 4
Question
Solve the following system of equations:
{x2+y2= 25
x+y= 7
3
Solution
Step 1: Rewrite the second equation in terms of one variable. Solve for either
xor yin terms of the other variable. We will solve for yin terms of x.
y= 7 −x
Step 2: Substitute the expression for yinto the first equation.
x2+ (7 −x)2= 25
Step 3: Expand and simplify the equation.
x2+ 49 −14x+x2= 25
2x2−14x+ 24 = 0
Step 4: Rearrange the equation into standard quadratic form.
x2−7x+ 12 = 0
Step 5: Factor the quadratic equation.
(x−3)(x−4) = 0
Step 6: Set each factor to zero and solve for x.
x= 3 or x= 4
Step 7: Substitute the values of xback into the expression we found for y
to find the corresponding values of y. For x= 3:y= 7 −3 = 4
For x= 4:y= 7 −4 = 3
Step 8: The solutions to the system of equations are (3,4) and (4,3).
Question 5
Question
Find the standard form of the equation of the ellipse with foci at (−2,0) and
(2,0) passing through the point (4,3).
Solution
Step 1: Find the center of the ellipse. To find the center of the ellipse, we
can take the average of the foci coordinates. The center of the ellipse is at the
midpoint of the foci: (−2+2
2,0+0
2)= (0,0).
Step 2: Find the distance between the foci. The distance between the foci
is 2(2) = 4 (since the foci have coordinates (±2,0)).
4
Step 3: Find the distance between one of the foci and the point (4,3). Using
the distance formula √(x2−x1)2+ (y2−y1)2, we have:
√(2 −4)2+ (0 −3)2=√(−2)2+ (−3)2=√4 + 9 = √13.
Step 4: Find the lengths of the semi-major axis (a) and the semi-minor axis
(b). The sum of the distances from the foci to any point on the ellipse is 2a
(the major axis length). Therefore, 2a= 4 =⇒a= 2. Also, a2=b2+c2
where cis the distance from the center to a focus. Since c= 2, we have
22=b2+ 22=⇒4 = b2+ 4 =⇒b2= 0 =⇒b= 0.
Step 5: Write the standard form of the equation of the ellipse. With the
center at the origin (0,0) and the values for aand b, the equation of the ellipse
is:
x2
4+y2
0= 1 .
Question 6
Question
Solve the equation 3x2+ 4xy +y2= 10 for xin terms of y.
Solution
Step 1: Rewrite the equation in standard form by completing the square for x.
3x2+ 4xy +y2= 10
3x2+ 4xy + 4 (y
2)2= 10 + 4 (y
2)2
3x2+ 4xy + 4 (y
2)2= 10 + 2y2
Step 2: Factor the trinomial.
(3x+ 2y)(x+ 2y) = 10 + 2y2
5
Step 3: Solve for xin terms of y.
(3x+ 2y)(x+ 2y) = 10 + 2y2
3x+ 2y=10 + 2y2
x+ 2y
3x+ 2y=10 + 2y2
x+ 2y
3x(x+ 2y)+2y(x+ 2y) = 10 + 2y2
3x2+ 6xy + 2xy + 4y2= 10 + 2y2
3x2+ 8xy + 4y2= 10 + 2y2
3x2+ 8xy + 4y2−2y2= 10
3x2+ 8xy + 2y2= 10
Step 4: Factor the trinomial.
(3x+y)(x+ 2y) = 10
Therefore, the solution to the equation 3x2+ 4xy +y2= 10 for xin terms
of yis x=10
3y+y.
Question 7
Question
Solve the system of equations:
{x2+y2= 25
x−y= 1
Solution
Step 1: Solve the second equation for x:
x=y+ 1
Step 2: Substitute the expression y+ 1 for xin the first equation:
(y+ 1)2+y2= 25
Step 3: Expand and simplify the equation:
y2+ 2y+1+y2= 25
6
2y2+ 2y−24 = 0
Step 4: Divide the equation by 2 to simplify:
y2+y−12 = 0
Step 5: Factor the quadratic equation:
(y+ 4)(y−3) = 0
Step 6: Set each factor to zero and solve for y:
y+ 4 = 0 =⇒y=−4
y−3 = 0 =⇒y= 3
Step 7: Use the values of yto find the corresponding values of x: For y=−4:
x=−4 + 1 =⇒x=−3
For y= 3:
x= 3 + 1 =⇒x= 4
Step 8: The solutions to the system of equations are:
(x, y) = (−3,−4),(4,3)
Question 8
Question
Solve the following system of equations:
{x2+y2= 25
x−y= 3
Solution
Step 1: Subtract the second equation from the first to eliminate y:
x2+y2−(x−y) = 25 −3
Step 2: Simplify the equation:
x2+y2−x+y= 22
Step 3: Simplify further by rearranging the terms:
x2−x+y2+y= 22
7
Step 4: Complete the square for both xand y: For x: Add (1/2)2= 1/4
inside the bracket:
x2−x+1
4+y2+y= 22 + 1
4
Step 5: Similarly, complete the square for y: Add (1/2)2= 1/4inside the
bracket:
x2−x+1
4+y2+y+1
4= 22 + 1
4+1
4
Step 6: Factor the left side:
(x−1
2)2
+(y+1
2)2
= 22 + 1
2
Step 7: Simplify the right side:
(x−1
2)2
+(y+1
2)2
=45
2
Step 8: Since the equation represents a circle, its center is at (1
2,−1
2)and
its radius is √45/2.
Thus, the solution to the system of equations is the point of intersection
between the circle x2+y2= 25 and the line x−y= 3.
Question 9
Question
Solve the following system of equations:
{x2+y2= 9
y=x2−2x
Solution
Step 1: Substitute the expression for yfrom the second equation into the first
equation.
x2+ (x2−2x)2= 9
Step 2: Simplify the equation.
x2+ (x4−4x3+ 4x2) = 9
Step 3: Combine like terms.
x4−4x3+ 5x2−9 = 0
Step 4: Factor the equation, if possible. In this case, factoring is not straight-
forward, so we’ll need to use numerical methods to find the solutions.
8
Step 5: Using a graphing calculator or numerical methods, we find that the
solutions to the equation are x=−1,x= 1,x≈1.78, and x≈2.78.
Step 6: Substitute these values back into the equation y=x2−2xto find
the corresponding yvalues.
Therefore, the solutions to the system of equations are (−1,3),(1,−1),
(1.78,−0.704), and (2.78,1.184).
Question 10
Question
Solve the following system of equations in terms of aand b:
{2x2−3y2=a
6x−5y=b
Solution
To solve this system of equations, we will use the substitution method to elim-
inate one of the variables.
Step 1: Solve the second equation for xin terms of y:We rearrange
the second equation to solve for x:
6x−5y=b
6x= 5y+b
x=5y+b
6
Step 2: Substitute xinto the first equation: Substitute the expression
for xinto the first equation of the system:
2(5y+b
6)2
−3y2=a
This simplifies to:
50y2+ 20by +b2
18 −3y2=a
Step 3: Simplify the equation and solve for y:Combine and simplify
the terms: 50y2+ 20by +b2
18 −54y2
18 =a
50y2+ 20by +b2−54y2= 18a
50y2−54y2+ 20by = 18a−b2
−4y2+ 20by = 18a−b2
9
−4y(y−5b) = 18a−b2
y(y−5b) = b2−18a
4
y2−5by =b2−18a
4
Step 4: Solve for xusing the obtained value for y:Substitute the
value of yback into the equation x=5y+b
6:
x=5(±√y2−5by) + b
6
x=5(±√b2
−18a
4) + b
6
Therefore, the solutions for xand yin terms of aand bare:
y=±√b2−18a
4, x =5(±√b2
−18a
4) + b
6
Question 11
Question
Solve the system of equations:
{4x2+y2= 16
x−2y= 3
Solution
To solve this system of equations, we will use substitution method.
Step 1: Solve the second equation for xin terms of y.
x= 3 + 2y
Step 2: Substitute xfrom the second equation into the first equation and
solve for y.
4(3 + 2y)2+y2= 16
4(9 + 12y+ 4y2) + y2= 16
36 + 48y+ 16y2+y2= 16
17y2+ 48y+ 20 = 0
10
(y+ 4)(17y+ 5) = 0
Therefore, y=−4or y=−5
17 .
Step 3: Substitute the values of yback into x= 3 + 2yto find the corre-
sponding values of x.
For y=−4, we get x= 3 + 2(−4) = 3 −8 = −5.
For y=−5
17 , we get x= 3 + 2 (−5
17 )= 3 −10
17 =51
17 −10
17 =41
17 .
Step 4: The solutions to the system of equations are x=−5, y =−4and
x=41
17 , y =−5
17 .
Question 12
Question
Find the standard form of the equation of a hyperbola given its foci at (−3,2)
and (3,2) and passing through the point (5,6).
Solution
Step 1: Find the center of the hyperbola.
The center of the hyperbola is the midpoint of the segment connecting the
foci. Use the midpoint formula:
(h, k) = (−3+3
2,2+2
2)
(h, k) = (0,2)
Step 2: Find the distance between the foci.
The distance between the foci of a hyperbola is 2a, where ais the distance
from the center to a vertex. Thus, we have:
√(3 −(−3))2+ (2 −2)2= 2a
6 = 2a
a= 3
Step 3: Find the equation of the hyperbola.
The standard form of the equation of a hyperbola centered at (h, k)with
vertices along the x-axis is:
(x−h)2
a2−(y−k)2
b2= 1
Since the hyperbola passes through the point (5,6), substitute x= 5 and y= 6
into the equation:
(5 −0)2
32−(6 −2)2
b2= 1
11
25/9−16/b2= 1
16/b2= 16/9
b2= 9
b= 3
Therefore, the standard form of the equation of the hyperbola is:
x2
9−(y−2)2
9= 1
Question 13
Question
Solve the following system of equations:
{x2+y2= 25
x+y= 7
Solution
Step 1: We can start by solving the second equation for xin terms of y:
x= 7 −y
Step 2: Substituting this expression for xinto the first equation gives:
(7 −y)2+y2= 25
Step 3: Simplifying the left side of the equation:
(49 −14y+y2) + y2= 25
49 −14y+ 2y2= 25
Step 4: Rearranging the equation:
2y2−14y+ 24 = 0
Step 5: Dividing the equation by 2 to simplify:
y2−7y+ 12 = 0
Step 6: Factoring the quadratic equation:
(y−4)(y−3) = 0
Step 7: Setting each factor to zero and solving for y:
y−4 = 0 =⇒y= 4
12
y−3 = 0 =⇒y= 3
Step 8: Now that we have the possible values for y, we can substitute them
back into the equation x= 7 −yto find the corresponding values of x: - For
y= 4:x= 7 −4 = 3 - For y= 3:x= 7 −3 = 4
Therefore, the solutions to the system of equations are (x, y) = (3,4) and
(x, y) = (4,3).
Question 14
Question
Solve the equation 3x2+ 4xy + 2y2−4x+ 4y+ 3 = 0 by classifying the conic
section it represents.
Solution
Step 1: To classify the conic section, we can rewrite the equation in matrix form
and find the eigenvalues of the resulting matrix. The matrix form of the given
equation is:
3 2 −2
2 2 2
−2 2 3
x
y
1
= 0
Step 2: To find the eigenvalues of the matrix, we set up and solve the
characteristic equation:
det
3−λ2−2
2 2 −λ2
−2 2 3 −λ
= 0
Step 3: Solving the characteristic equation, we find the eigenvalues λ1= 2,
λ2= 5, and λ3= 1.
Step 4: Since we have both positive and negative eigenvalues, the conic
section represented by the given equation is a hyperbola.
Question 15
Question
Solve the following system of equations:
{2x2+y2= 18
x2−2y2= 2
13
Solution
Step 1: Start by isolating y2in the first equation.
{y2= 18 −2x2
x2−2y2= 2
Step 2: Substitute y2= 18 −2x2into the second equation.
x2−2(18 −2x2) = 2
Step 3: Simplify and solve for x.
x2−36 + 4x2= 2
5x2= 38
x2=38
5
x=±√38
5
Step 4: Substitute xback into y2= 18 −2x2to find y.
y2= 18 −2(38
5)
y2= 18 −76
5
y2=90
5−76
5
y2=14
5
y=±√14
5
Step 5: The solutions to the system of equations are:
(√38
5,√14
5)and (−√38
5,−√14
5)
Question 16
Question
Solve the following system of equations:
{x2+ 2y2= 2
4x2−y2= 5
14
Solution
Step 1: First, we can rearrange the second equation to solve for y2:
4x2−y2= 5 =⇒y2= 4x2−5
Step 2: Substitute the expression for y2into the first equation:
x2+ 2(4x2−5) = 2
Step 3: Simplify the equation:
x2+ 8x2−10 = 2 =⇒9x2= 12 =⇒x2=4
3
Step 4: Take the square root of both sides to solve for x:
x=±√4
3=±2
√3=±2√3
3
Step 5: Substitute the values of xback into one of the original equations.
We’ll use the first equation:
{(2√3
3)2+ 2y2= 2
(7√3)2−y2= 5
Step 6: Simplify the equations:
{4
3+ 2y2= 2
21 −y2= 5
Step 7: Solve for yin the second equation:
y2= 21 −5 = 16 =⇒y=±4
Step 8: Solve for yusing y=±4in the first equation:
{4
3+ 2(4)2= 2 or 4
3+ 2(−4)2= 2
21 −(4)2= 5 or 21 −(−4)2= 5
Step 9: Simplify and solve for y:
{4
3+ 32 = 2 or 4
3+ 32 = 2
21 −16 = 5 or 21 −16 = 5
Step 10: Check our solutions: Both sets of solution should satisfy the original
system of equations:
{(2√3
3)2+ 2(4)2= 2
(7√3)2−42= 5 and {(−2√3
3)2+ 2(−4)2= 2
(−7√3)2−42= 5
Hence, the solutions are x=±2√3
3and y=±4.
15
Question 17
Question
Solve the following system of equations:
{x2+y2= 25
x−y= 1
Solution
Step 1: Solve the second equation for either variable. Let’s solve for xin terms
of y:
x=y+ 1
Step 2: Substitute xin terms of yinto the first equation:
(y+ 1)2+y2= 25
Step 3: Expand and simplify the equation:
y2+ 2y+1+y2= 25
2y2+ 2y−24 = 0
Step 4: Divide by 2 to simplify the equation:
y2+y−12 = 0
Step 5: Factor the quadratic equation:
(y+ 4)(y−3) = 0
Step 6: Set each factor to zero and solve for y:
y+ 4 = 0 =⇒y=−4
y−3 = 0 =⇒y= 3
Step 7: Substitute the values of yback into the equation x=y+ 1 to find
the corresponding values of x:
For y=−4 : x=−4 + 1 = −3
For y= 3 : x= 3 + 1 = 4
Step 8: Therefore, the solutions to the system of equations are:
{x=−3, y =−4
x= 4, y = 3
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Question 18
Question
Solve the following system of equations:
{y=x2+ 4x
y= 3x+ 6
Solution
Step 1: We start by setting the two expressions for yequal to each other to find
the value(s) of x:
x2+ 4x= 3x+ 6
Step 2: Rearrange the equation and set it equal to zero by subtracting 3x
and 6from both sides:
x2+ 4x−3x−6 = 0
Step 3: Simplify the equation:
x2+x−6 = 0
Step 4: Factor the quadratic equation:
(x−2)(x+ 3) = 0
Step 5: Set each factor to zero and solve for x:
x−2 = 0 or x+ 3 = 0
Step 6: Solve for x:
x= 2 or x=−3
Step 7: Substitute the value of xback into the equation y=x2+ 4xto find
the corresponding values of y: For x= 2:
y= 22+ 4(2) = 4 + 8 = 12
So, when x= 2,y= 12.
For x=−3:
y= (−3)2+ 4(−3) = 9 −12 = −3
So, when x=−3,y=−3.
Step 8: The solution to the system of equations is therefore:
(x, y) = (2,12),(−3,−3)
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Question 19
Question
Consider the equation of the hyperbola (x−1)2
9−(y+2)2
4= 1. Determine the
vertices, foci, and equations of the asymptotes of this hyperbola.
Solution
Step 1: First, identify key information from the given equation. The general
equation of a hyperbola centered at (h, k)is of the form (x−h)2
a2−(y−k)2
b2= 1
for a hyperbola opening horizontally or (y−k)2
a2−(x−h)2
b2= 1 for a hyperbola
opening vertically. In our case, the center of the hyperbola is at (1,−2),a2= 9,
and b2= 4.
Step 2: The vertices of the hyperbola are given by the points (h±a, k)for a
horizontally-opening hyperbola and (h, k±a)for a vertically-opening hyperbola.
So, the vertices are at (1 + 3,−2) = (4,−2) and (1 −3,−2) = (−2,−2).
Step 3: The distance from the center to the foci is given by c=√a2+b2.
In this case, c=√9 + 4 = √13. The foci are located at a distance of √13 units
from the center along the transverse axis, in this case, the x-axis. Therefore,
the foci are at (1 + √13,−2) and (1 −√13,−2).
Step 4: The slopes of the asymptotes of a hyperbola are given by ±b
a. In
this case, the slopes are ±2
3. Since the center of the hyperbola is at (1,−2), the
asymptotes will have the equation y+ 2 = ±2
3(x−1). Simplifying, we get the
equations of the asymptotes as y=2
3x−8
3and y=−2
3x−4
3.
Question 20
Question
Consider the conic section defined by the equation x2
4+y2
9−2x+ 6y+ 1 = 0.
Find the standard form of this conic section, determine its type, and sketch its
graph.
Solution
Step 1: Rearrange the equation into standard form by completing the square
for xand y:
x2
4+y2
9−2x+ 6y+ 1 = 0
x2
4−2x+y2
9+ 6y=−1
x2
4−2x+1+y2
9+ 6y+ 9 = −1+1+9
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(x
2−1)2+(y
3+ 2)2= 9
Step 2: The standard form of the conic section is (x
2−1)2+(y
3+ 2)2= 32.
Step 3: Since the coefficients of x2and y2are positive and equal, the conic
section is a circle.
Step 4: To sketch the graph, we note that the center of the circle is at (2,−2)
and the radius is 3. Therefore, plot the point (2,−2) as the center and draw a
circle of radius 3.
Step 5: Sketch the circle on the coordinate plane.
Thus, the standard form of the conic section is (x
2−1)2+(y
3+ 2)2= 9, it
is a circle, and the graph is centered at (2, -2) with a radius of 3.
Question 21
Question
Solve the system of equations:
{x2+ 4y2= 16
x2−2x+ 2y2−4y= 7
Solution
Step 1: Begin by rewriting the second equation in the system by completing the
square for both the xand yterms:
x2+ 4y2= 16
(x2−2x) + (2y2−4y) = 7
(x2−2x+ 1) + (2(y2−2y+ 1)) = 7 + 1 + 2
(x−1)2+ 2(y−1)2= 10
Step 2: Make a substitution by letting X=x−1and Y=y−1so that the
equation becomes:
{x2+ 4y2= 16
X2+ 2Y2= 10
Step 3: Multiply the second equation by 4:
{x2+ 4y2= 16
4X2+ 8Y2= 40
Step 4: Add the first equation to the multiple of the second equation:
5X2= 56 =⇒X2=56
5
19
Step 5: Solve for X:
X=±√56
5=±2√14
5
Step 6: Substitute the value of Xback in and solve for Y:
2Y2= 10 −4( 14
5) = 10 −11.2 = −1.2 =⇒Y2=−0.6
Step 7: Since Y2is negative, there are no real solutions to this system of
equations.
Question 22
Question
Solve the system of equations:
{x2−y2= 16
xy =−8
Solution
Step 1: Let’s solve the second equation for yin terms of x.
xy =−8 =⇒y=−8
x
Step 2: Substitute y=−8
xinto the first equation.
x2−(−8
x)2
= 16
Step 3: Simplify the equation by squaring and combining terms.
x2−64
x2= 16
Step 4: Multiply through by x2to clear the fraction.
x4−64 = 16x2
Step 5: Rearrange the equation into standard form and solve for x.
x4−16x2−64 = 0
Step 6: Let u=x2to simplify the equation.
u2−16u−64 = 0
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Step 7: Factor the quadratic equation.
(u−8)(u+ 8) = 0
Step 8: Set each factor to zero and solve for u.
u= 8 or u=−8
Step 9: Substitute back u=x2to find possible values for x. For u= 8:
x2= 8 =⇒x=±2√2. For u=−8: No real solutions.
Step 10: Finally, substitute the values of xback into y=−8
xto find the
corresponding yvalues. Thus, the solutions to the system of equations are:
(x, y) = (2√2,−4√2) and (−2√2,4√2) .
Question 23
Question
Solve the following system of equations:
{4x2+ 9y2= 36
x2−y2= 1
Solution
Step 1: We can start by solving the second equation for y2:
x2−y2= 1
y2=x2−1
Step 2: Now we can substitute y2=x2−1into the first equation:
4x2+ 9(x2−1) = 36
Step 3: Simplify the equation:
4x2+ 9x2−9 = 36
13x2−9 = 36
Step 4: Add 9 to both sides:
13x2= 45
Step 5: Divide by 13:
x2=45
13
21
Step 6: Taking the square root of both sides, we get two possible solutions
for x:
x=±√45
13
Step 7: Substituting back into y2=x2−1to find y: For x=√45
13 :
y2=45
13 −1 = 32
13
y=±√32
13
For x=−√45
13 :
y2=45
13 −1 = 32
13
y=±√32
13
Therefore, the solutions to the system of equations are:
(x, y) = (√45
13,±√32
13)and (−√45
13,±√32
13)
Question 24
Question
Let A(1,3),B(5,6), and C(2,−2) be three points in the plane. Determine the
equation of the circle passing through these three points.
Solution
Step 1: Find the equations of the perpendicular bisectors of the line segments
AB and BC.
To find the equation of the perpendicular bisector of the line segment AB,
we first find the midpoint Mof AB, which is given by:
M(1+5
2,3+6
2)=M(3,4.5).
The slope of the line passing through Aand Bis:
mAB =6−3
5−1=3
4.
22
Therefore, the slope of the perpendicular bisector of AB is −4
3. Since this line
passes through the midpoint M(3,4.5), its equation can be written as:
y−4.5 = −4
3(x−3).
Similarly, for the line passing through Band C, first find the midpoint N
of BC, which is given by:
N(5+2
2,6+(−2)
2)=N(7
2,2).
The slope of the line passing through Band Cis:
mBC =2−6
2−5=−4
3.
Thus, the slope of the perpendicular bisector of BC is 3
4. Hence, its equation
is:
y−2 = 3
4(x−7
2).
Step 2: Solve the system of equations formed by the perpendicular bisectors
calculated in Step 1.
Solve the system of equations
{y−4.5 = −4
3(x−3)
y−2 = 3
4(x−7
2)
to find the point Oat which the perpendicular bisectors intersect.
Step 3: Determine the radius of the circle.
Calculate the distance between the point Oand any of the given points A,
B, or Cto find the radius of the circle.
Step 4: Write the equation of the circle.
Finally, using the center Oand the radius, write the equation of the circle
passing through points A,B, and C.
Question 25
Question
Find the standard form of the equation of the hyperbola with vertices at (−5,0)
and (5,0), and foci at (−7,0) and (7,0).
Solution
Step 1: Find the center of the hyperbola. The center of the hyperbola is the
midpoint of the segment connecting the vertices. The midpoint formula is given
by (x1+x2
2,y1+y2
2). Therefore, the center is ((5 + (−5))/2,(0 + 0)/2) = (0,0).
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Step 2: Find the distance from the center to a vertex. This distance is called
the distance from the center to the vertex, which is a= 5 −0 = 5.
Step 3: Find the distance from the center to a focus. This distance is called
the distance from the center to the focus, which is c= 7 −0 = 7.
Step 4: Find the value of busing the relationship c2=a2+b2. Substitute
a= 5 and c= 7 into the equation and solve for b:
72= 52+b2
49 = 25 + b2
b2= 24
b=√24 = 2√6
Step 5: Write the equation of the hyperbola using the given information.
The standard form of the equation of a hyperbola centered at the origin is
x2
a2−y2
b2= 1. Substituting the values a= 5 and b= 2√6into the equation, we
have: x2
52−y2
(2√6)2= 1
x2
25 −y2
24 = 1
Therefore, the standard form of the equation of the hyperbola is x2
25 −y2
24 = 1.
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