MATH 121 - COLLEGE ALGEBRA Equations and properties of conic sections Question Bank Set 7

MATH 121 - COLLEGE ALGEBRA -

Equations and properties of conic

sections

Question Bank - Set 7

Liberty University

Question 1

Question

Find the standard form of the equation of the parabola with focus at (3,−2)

and directrix y= 5.

Solution

Step 1: First, we determine whether the parabola opens horizontally or verti-

cally. Since the directrix is y= 5, which is a horizontal line, the parabola opens

horizontally.

Step 2: We can denote the coordinates of the vertex of the parabola as

(h, k). The distance between the vertex and the focus is the same as the distance

between the vertex and the directrix. Therefore, the distance between (h, k)and

(3,−2) (the focus) is equal to the distance between (h, k)and the line y= 5

(the directrix).

Step 3: Using the distance formula, we have:

(h−3)2+ (k+ 2)2= (k−5)2

Step 4: To simplify, we expand and consolidate terms to obtain:

h2−6h+9+k2+ 4k+ 4 = k2−10k+ 25

Step 5: Further simplifying gives:

h2−6h+k2+ 4k+ 13 = −10k

Step 6: Rearrange the equation to standard form by combining like terms:

h2−6h+k2+ 4k+ 10k+ 13 = 0

Step 7: Completing the square in terms of hand krespectively, we get:

(h−3)2+ (k+ 2)2= 0

Therefore, the standard form of the equation of the parabola is (y+ 2)2= 0 .

Question 2

Question

Solve the equation x2−y2+ 4x+ 6y−12 = 0 by completing the square, and

determine the type of conic section represented by the equation.

Solution

Step 1: Rearrange the equation by grouping the xand yterms together:

(x2+ 4x)−(y2−6y) = 12

Step 2: Complete the square for the xterms: 1. Take half of the coeﬀicient

of x, square it, and add/subtract it inside the parentheses:

(x2+ 4x+ 4) −(y2−6y) = 12 + 4

2. Simplify the square and the constant term:

(x+ 2)2−(y2−6y) = 16

Step 3: Complete the square for the yterms: 1. Take half of the coeﬀicient

of y, square it, and add/subtract it inside the parentheses:

(x+ 2)2−(y2−6y+ 9) = 16 + 9

2. Simplify the square and the constant term:

(x+ 2)2−(y−3)2= 25

Step 4: Compare the equation with the standard form of conic sections.

Since the signs of the squared terms are different, the equation represents a

hyperbola.

Question 3

Question

Let Cbe the conic section defined by the equation 4x2+ 9y2= 36. Find the

standard form of C.

Solution

Step 1: Divide both sides of the equation by 36 to get the standard form. Step

2: Rearrange the equation to isolate the terms.

Step 1:

4x2

36 +9y2

36 = 1

9+y2

4= 1

Step 2:

9+y2

4= 1

32−y2

22= 1

(x−0)2

32−(y−0)2

22= 1

Therefore, the standard form of the conic section Cis (x−0)2

32−(y−0)2

22= 1.

Question 4

Question

Solve the following system of equations:

{x2+y2= 25

x2−y2= 9

Solution

Step 1: We’ll start by solving the second equation in the system for x2in terms

of y.

x2= 9 + y2

Step 2: Substitute this expression for x2into the first equation of the system.

9 + y2+y2= 25

Step 3: Simplify the equation.

2y2= 16

Step 4: Solve for y.

y2= 8 =⇒y=±√8 = ±2√2

Step 5: Substitute the values of yback into x2= 9 + y2to find the corre-

sponding values of x. For y= 2√2,

x2= 9 + (2√2)2= 9 + 8 = 17 =⇒x=±√17

For y=−2√2,

x2= 9 + (−2√2)2= 9 + 8 = 17 =⇒x=±√17

Thus, the solutions to the system are:

(x, y) = (√17,2√2),(−√17,2√2),(√17,−2√2),(−√17,−2√2)

Question 5

Question

Find the standard form of the equation of the ellipse with foci at (−4,0) and

(4,0), and a major axis of length 8.

Solution

Step 1: The distance between the foci is equal to 2a, where ais the distance

from the center of the ellipse to a vertex along the major axis. In this case,

2a= 8 =⇒a= 4.

Step 2: The center of the ellipse is the midpoint of the line segment joining

the foci. The midpoint of the line segment with endpoints (−4,0) and (4,0) is

(0,0), so the center of the ellipse is at (0,0).

Step 3: The distance from the center of the ellipse to each focus is c. Since

cis 4 units from the center to each focus, then c= 4.

Step 4: The equation for an ellipse with center at the origin is x2

a2+y2

b2= 1,

where ais the distance from the center to a vertex along the major axis, and b

is the distance from the center to a vertex along the minor axis. Our equation

will be x2

42+y2

b2= 1.

Step 5: To find b, we use the relationship a2=b2+c2. Substituting the

values of aand c, we get 42=b2+42=⇒16 = b2+16 =⇒b2= 0 =⇒b= 0.

Step 6: Therefore, the equation for the ellipse is x2

16 +y2

0= 1, which simplifies

to x2

16 = 1 or x2= 16.

Step 7: The standard form of the equation of the ellipse is x2= 16.

Question 6

Question

Consider the conic section represented by the equation 4x2+y2−8x+6y+16 = 0.

Determine the type of conic section, find the center, vertices, and foci, and sketch

the graph of the conic section.

Solution

Step 1: Rewrite the given equation in standard form by completing the square

for both xand yterms.

4x2+y2−8x+ 6y+ 16 = 0

4(x2−2x) + y2+ 6y=−16

4(x2−2x+ 1) + y2+ 6y+ 9 = −16+4+9

4(x−1)2+ (y+ 3)2=−3

Step 2: Determine the type of conic section by examining the coeﬀicients of

the x2and y2terms. Since both coeﬀicients are positive and unequal, the conic

section is an ellipse.

Step 3: Find the center of the ellipse by looking at the completed square

form. The center of the ellipse is at (h, k) = (1,−3).

Step 4: Compute the vertices of the ellipse. The vertices are located aunits

to the left and right of the center, where ais the square root of the denominator

in the xterm. In this case, a=√3. Therefore, the vertices are at (1 −√3,−3)

and (1 + √3,−3).

Step 5: Calculate the distance from the center to the foci. The distance

is denoted by cand is found using the formula c=√a2−b2, where bis the

square root of the denominator in the yterm. Here, b=√3and a=√3. Thus,

c=√3.

Step 6: Locate the foci. The foci are situated cunits from the center along

the major axis. Hence, the foci are at (1 + √3,−3) and (1 −√3,−3).

Step 7: Sketch the ellipse by plotting the center, vertices, and foci on a

coordinate plane. Draw the ellipse passing through the vertices with the foci

located inside the ellipse.

Question 7

Question

Find the standard form of the equation of the parabola with focus F(0,3) and

directrix y=−3.

Solution

Step 1: First, we need to determine whether the parabola opens vertically or

horizontally. Since the directrix is parallel to the x-axis, the parabola opens

vertically.

Step 2: The distance between the vertex (h, k)of the parabola and the focus

is equal to the distance between the vertex and the directrix. Given that the

focus is F(0,3), and directrix is y=−3, the vertex has coordinates (0,0).

Step 3: Since the parabola opens vertically, the general form of the equation

of a parabola is (x−h)2= 4p(y−k). Since the vertex is (h, k) = (0,0), the

equation simplifies to x2= 4py.

Step 4: The distance between the vertex and the focus is the same as the

distance between the vertex and the directrix, which is |p|= 3. Therefore,

p= 3.

Step 5: Substitute the value of pinto the equation x2= 4py to obtain the

standard form of the equation. So, the equation of the parabola is x2= 12y.

Question 8

Question

Solve the following system of equations:

{x2+y2= 25

3x−4y= 6

Solution

Step 1: Let’s solve the second equation for xin terms of y:

3x−4y= 6

3x= 4y+ 6

x=4

3y+ 2

Step 2: Now substitute this expression for xinto the first equation:

3y+ 2)2

+y2= 25

9y2+16

3y+4+y2= 25

9y2+16

3y−21 = 0

Step 3: Multiply the equation by 9 to get rid of the fraction:

25y2+ 48y−189 = 0

Step 4: Now we have a quadratic equation. To solve for y, we can use the

quadratic formula:

y=−b±√b2−4ac

where a= 25,b= 48, and c=−189.

Step 5: Plugging in the values, we get:

y=−48 ±√482−4∗25 ∗(−189)

2∗25

y=−48 ±√2304 + 18900

y=−48 ±√21204

y=−48 ±2√5301

Step 6: This gives us two potential solutions for y. Now plug each value of

yback into the equation x=4

3y+ 2 to find the corresponding values of x.

Step 7: Finally, the solutions to the system of equations are the pairs of

(x, y)values that satisfy both equations simultaneously.

Question 9

Question

Let y=−2x2+ 8 be the equation of a parabola. Determine the coordinates of

the vertex, the focus, and the directrix of the parabola.

Solution

Step 1: We first rewrite the equation in standard form by completing the square

for x. Step 2: Rearrange the equation to center the parabola. Step 3: Compare

the new equation to the standard form of a parabola to identify the vertex,

the focus, and the directrix. Step 4: Write down the final answers for the

coordinates of the vertex, the focus, and the directrix.

Question 10

Question

A parabola has a vertex at the point (−2,3) and a focus at the point (4,3).

Determine the equation of the parabola in standard form.

Solution

Step 1: Recall that for a parabola with a horizontal axis of symmetry, the

equation in standard form is (x−h)2= 4p(y−k), where (h, k)is the vertex

and pis the distance from the vertex to the focus (or directrix).

Step 2: First, we identify the vertex as (h, k) = (−2,3).

Step 3: Next, we find the distance from the vertex to the focus. Since the

focus is at (4,3), the distance is p= 4 −(−2) = 6.

Step 4: Substitute the vertex and pinto the standard form equation to get

(x+ 2)2= 24(y−3).

Step 5: Simplify the equation to get the final answer in standard form:

(x+ 2)2= 24y−72.

Question 11

Question

Solve the equation 3x2−2y2+ 18x+ 4y−22 = 0 for y.

Solution

Step 1: Rewrite the given equation in standard form by completing the square

for both xand yterms.

3x2−2y2+ 18x+ 4y−22 = 0

3(x2+ 6x)−2(y2−2y) = 22

3(x2+ 6x+ 9) −2(y2−2y+ 1) = 22 + 3(9) −2(1)

3(x+ 3)2−2(y−1)2= 49

Step 2: Divide both sides of the equation by 49 to obtain 1on the right side.

3(x+ 3)2

49 −2(y−1)2

49 = 1

(x+ 3)2

(7)2−(y−1)2

(7/√2)2= 1

Step 3: Compare the equation obtained with the standard form of hyperbola

to find values of aand b. The equation of a hyperbola in standard form is

(x−h)2

a2−(y−k)2

b2= 1. Comparing it with the given equation, we have a= 7 and

b=7

√2.

Step 4: The standard form of a hyperbola centered at (h, k)with major axis

along the x-axis is (x−h)2

a2−(y−k)2

b2= 1. Thus, the equation of hyperbola in

standard form for this problem is (x+3)2

(7)2−(y−1)2

(7/√2)2= 1.

Question 12

Question

Solve the following system of equations:

{x2+ 4y2= 5

x2−y2= 1

Solution

Step 1: Start by isolating x2in the second equation:

x2=y2+ 1

Step 2: Substitute x2in terms of y2from the second equation into the first

equation:

(y2+1)+4y2= 5

Step 3: Simplify the equation by combining like terms:

5y2+ 1 = 5

Step 4: Subtract 1 from both sides of the equation:

5y2= 4

Step 5: Divide by 5 to solve for y2:

y2=4

Step 6: Take the square root of both sides to solve for y:

y=±√4

Step 7: Simplify the square root to obtain the values of y:

y=±2

√5

Step 8: Substitute the values of yback into the equation x2=y2+1 to solve

for x:

x2=(2

√5)2

+ 1 or x2=(−2

√5)2

+ 1

Step 9: Simplify to find the values of x:

x=±√1

5+ 1 or x=±√1

5+ 1

Step 10: Simplify further to get the final solutions for x:

x=±√6

5or x=±√6

Therefore, the solutions to the system of equations are:

(x, y) = (√6

5,2

√5),(−√6

5,−2

√5),(√6

5,−2

√5),(−√6

5,2

√5)

Question 13

Question

Solve the system of equations for xand y:

{3x2−4xy +y2= 10

2x2+ 3xy −5y2=−7

Solution

Step 1: Multiply the first equation by 5 and the second equation by 2 to make

the coeﬀicient of y2the same in both equations:

{15x2−20xy + 5y2= 50

4x2+ 6xy −10y2=−14

Step 2: Add the two equations together to eliminate the term y2:

19x2−14y2= 36

Step 3: This equation represents a hyperbola. To simplify, divide by 36:

(√36

19 )2−y2

(√36

14 )2= 1

6√19

19 −y2

6√14

= 1

6√19

19 −y2

3√14

= 1

Step 4: Comparing this equation with the standard form of a hyperbola,

a2−y2

b2= 1, we see that a=6√19

19 and b=3√14

Therefore, the solution to the system of equations is a hyperbola with a=

6√19

19 and b=3√14

Question 14

Question

Find the standard form of the equation of the hyperbola that satisfies the given

conditions: foci at (±5,0) and passes through the point (4,3).

Solution

Step 1: Determine the center of the hyperbola. Since the foci are on the x-axis,

the center of the hyperbola will also lie on the x-axis. The x-coordinate of the

center is the average of the x-coordinates of the foci:

Center =(−5+5

2,0)= (0,0)

Step 2: Write the equation of the hyperbola in standard form. The general

form of the equation of a hyperbola centered at (h, k)is (x−h)2

a2−(y−k)2

b2= 1.

We need to find the values of aand b.

Step 3: Determine the distance from the center to one of the foci, which is

the value of c. In this case, c= 5.

Step 4: Use the distance formula to determine the equation of the ellipse.

The distance between the point (4,3) and a focus (5,0) is equal to 2a(the

distance from the center to a vertex). Therefore,

√(4 −5)2+ (3 −0)2= 2a

√(−1)2+ 32= 2a

√10 = 2a

a=√10

Step 5: Find the value of busing a,c, and the relationship between a,b, and

cin hyperbolas:

c2=a2+b2

52=(√10

2)2

+b2

25 = 10

4+b2

b2= 25 −10

b2=90

b=3√10

Step 6: Write the equation of the hyperbola in standard form.

(√10/2)2−y2

(3√10/2)2= 1

5/2−y2

45/2= 1

2x2−2y2

9= 1

Question 15

Question

Solve the following system of equations:

{x2+y2= 25

x−y= 3

Solution

Step 1: We start by solving the second equation for xin terms of y.

x−y= 3

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+ 9 = 25

2y2+ 6y−16 = 0

Step 4: Solve the quadratic equation 2y2+ 6y−16 = 0 using the quadratic

formula.

y=−b±√b2−4ac

Step 5: Plug in a= 2,b= 6, and c=−16 into the quadratic formula.

y=−6±√62−4(2)(−16)

2(2)

y=−6±√36 + 128

y=−6±√164

Step 6: Simplify the expression under the square root.

√164 = √4×41 = 2√41

Step 7: Substitute this back into the equation for y.

y=−6±2√41

y=−3±√41

Step 8: Solve for xusing x=y+ 3.

x=−3±√41

2+ 3

x=−3±√41

2+6

x=3±√41

Step 9: So, the solutions to the system of equations are:

{x=3+√41

2, y =−3+√41

x=3−√41

2, y =−3−√41

Question 16

Question

Determine the standard form of the equation of the parabola with focus at

F(3,4) and directrix x= 1.

Solution

Step 1: Recall the definition of a parabola: The set of all points that are

equidistant from a fixed point (the focus) and a fixed line (the directrix).

Step 2: Determine the vertex of the parabola. Since the focus is at F(3,4)

and the directrix is x= 1, the vertex lies halfway between the focus and the

directrix. Thus, the vertex is at V(Fx+D

2,Fy

2), where Dis the x-coordinate of

the directrix. Therefore, the vertex is V(3+1

2,4

2)= (2,2).

Step 3: Determine the distance between the focus and the vertex. This

distance is called the focal length, denoted by |p|. Since the focus is at F(3,4)

and the vertex is at (2,2),|p|=|Fx−Vx|=|3−2|= 1.

Step 4: Determine the equation of the parabola. Since the directrix is vertical

and to the left of the vertex, the parabola opens to the right. The standard form

of the equation of the parabola in this case is (x−h)2= 4p(y−k), where (h, k)

is the vertex.

Step 5: Substitute the known values into the standard form equation: (x−

2)2= 4(1)(y−2).

Step 6: Simplify the equation: (x−2)2= 4(y−2).

Therefore, the standard form of the equation of the parabola is (x−2)2=

4(y−2).

Question 17

Question

Solve the equation for y:4x2+ 9y2+ 8x−36y+ 23 = 0.

Solution

Step 1: Rearrange the equation by grouping the xterms and completing the

square for x.

4x2+ 9y2+ 8x−36y+ 23 = 0

4x2+ 8x+ 9y2−36y=−23

4(x2+ 2x) + 9(y2−4y) = −23

4(x2+ 2x+ 1) + 9(y2−4y+ 4) = −23 + 4(1) + 9(4)

4(x+ 1)2+ 9(y−2)2= 10

Step 2: Divide by 10 to get equation in standard form.

4(x+ 1)2

10 +9(y−2)2

10 = 1

(x+ 1)2

+(y−2)2

= 1

Step 3: Identify the ellipse’s center, vertices, foci, and eccentricity. The

equation (x−h)2

a2+(y−k)2

b2= 1 represents an ellipse centered at (h, k)with semi-

major axis aand semi-minor axis b. Comparing with our equation, we find:

(h, k) = (−1,2)

a=√10

4=√10

b=√10

9=√10

The center of the ellipse is (−1,2), semi-major axis is √10

2, and semi-minor

axis is √10

3. Hence, the ellipse’s vertices are (−1±√10

2,2), foci are (−1±√6,2),

and eccentricity is e=√1−b2

a2=√1−10

4·9=√14

36 =√7

Question 18

Question

Solve the equation 3x2+ 4y2+ 6x−16y+ 9 = 0 by completing the square.

Identify the conic section represented by the equation.

Solution

Step 1: Rewrite the equation by grouping xterms together and yterms together:

(3x2+ 6x) + (4y2−16y) + 9 = 0

Step 2: Complete the square for the xand yterms separately. For the x

terms:

(3x2+ 6x)

3(x2+ 2x)

3(x2+ 2x+ 1 −1)

3[(x+ 1)2−1]

3(x+ 1)2−3

For the yterms:

(4y2−16y)

4(y2−4y)

4(y2−4y+ 4 −4)

4[(y−2)2−4]

4(y−2)2−16

Adding the completed square expressions together, we have:

3(x+ 1)2−3 + 4(y−2)2−16 + 9 = 0

Step 3: Combine like terms:

3(x+ 1)2+ 4(y−2)2−10 = 0

Step 4: Divide the entire equation by 10 to simplify:

10(x+ 1)2+4

10(y−2)2−1 = 0

Step 5: Identify the conic section represented by the equation. Since the

coeﬀicients of (x+1)2and (y−2)2have the same sign, and the larger coeﬀicient

is on the xterm, the conic section is an ellipse.

Question 19

Question

Solve the system of equations:

{2x2+y2= 16

x2−2y2= 3

Solution

Step 1: Let’s first rewrite the system of equations in standard form by expanding

the terms: {2x2+y2= 16 (1)

x2−2y2= 3 (2)

Step 2: To eliminate a variable, let’s multiply equation (2) by 2:

2(x2−2y2) = 2(3)

2x2−4y2= 6

Step 3: Subtract equation (1) from the new equation we obtained:

(2x2−4y2)−(2x2+y2) = 6 −16

−5y2=−10

Step 4: Solve for y2:

y2= 2

Step 5: Substitute y2= 2 back into equation (1):

2x2+ 2 = 16

2x2= 14

x2= 7

Step 6: Take the square root of both sides to solve for x:

x=±√7

Therefore, the solution to the system of equations is:

(x, y) = (√7,√2),(−√7,√2)

Question 20

Question

Solve the system of equations:

{x2+y2−4x+ 6y+ 4 = 0

3x−4y+ 2 = 0

Solution

Step 1: We can rewrite the first equation in the standard form of a circle by

completing the square for xand yterms:

x2+y2−4x+ 6y+ 4 = 0

(x2−4x)+(y2+ 6y) = −4

(x2−4x+4)+(y2+ 6y+ 9) = −4+4+9

(x−2)2+ (y+ 3)2= 9

Step 2: The equation (x−2)2+ (y+ 3)2= 9 represents a circle with center

at (2,−3) and radius 3.

Step 3: Now, we substitute 3x−4y+ 2 = 0 into the equation of the circle:

3x−4y+ 2 = 0

3x= 4y−2

x=4

3y−2

Substituting xback into the equation of the circle, we get:

3y−2

3−2)2+ (y+ 3)2= 9

Step 4: Solving for yin the equation above may involve some algebraic

manipulations to simplify the expression and solve for y.

y=−2or y=−4

Step 5: Finally, substituting the values of yback into x=4

3y−2

3, we find

the corresponding values of xfor each yvalue.

Therefore, the solutions to the system of equations are:

(x, y) = (2

3,−2)and (x, y) = (2

3,−4)

Question 21

Question

Find the equation of the hyperbola with foci at (−2,1) and (4,1) and vertices

at (−5,1) and (7,1).

Solution

Step 1: Find the center of the hyperbola. The center of a hyperbola is the mid-

point between its vertices, so we can find the center by averaging the coordinates

of the vertices. The center is ((−5 + 7)/2,1) = (1,1).

Step 2: Find the distance between the center and one of the foci. This

distance is equal to c, the distance from the center to the focus. c= 4 −1 = 3.

Step 3: Find the distance between the center and one of the vertices. This

distance is equal to a, the distance from the center to the vertex. a= 7 −1 = 6.

Step 4: Use the relationship for hyperbolas: c2=a2+b2where ais the

distance from the center to the vertex and bis half the length of the transverse

axis. We can solve for b.32= 62+b29 = 36 + b2b2=−27 b=√−27 b=i√27

Step 5: Since our hyperbola is centered at (1,1) and the transverse axis is

horizontal, the equation has the form (x−1)2

62−(y−1)2

(i√27)2= 1. Simplifying, we get

(x−1)2

36 +(y−1)2

27 = 1 as the equation of the hyperbola.

Question 22

Question

Solve the following system of equations:

{4x2+ 9y2= 36

2x−3y= 1

Solution

Step 1: Solve the second equation for xin terms of y:

2x−3y= 1

2x= 3y+ 1

x=3y+ 1

Step 2: Plug this expression for xinto the first equation:

4(3y+ 1

2)2

+ 9y2= 36

4·9y2+ 6y+ 1

4+ 9y2= 36

Step 3: Simplify the equation:

9y2+ 6y+1+9y2= 36

18y2+ 6y+ 1 = 36

18y2+ 6y−35 = 0

Step 4: Solve the quadratic equation for yusing the quadratic formula:

y=−b±√b2−4ac

y=−6±√62−4(18)(−35)

2(18)

Step 5: Calculate the values of y:

y=−6±√36 + 2520

y=−6±√2556

y=−6±2√639

y=−3±√639

Step 6: Now that we have the values for y, we can substitute them back

into the equation x=3y+1

2to find the corresponding values of x. Thus, the

solutions to the system of equations are the pairs (x, y).

Question 23

Question

Consider the conic section given by the equation 3x2+4xy−4y2−14x+14y+13 =

0. Determine the type of conic section and find the equation of its axis.

Solution

Step 1: Identify the type of conic section To determine the type of conic

section, we will analyze the discriminant of the general conic equation Ax2+

Bxy +Cy2+Dx +Ey +F= 0 given by B2−4AC.

For the equation 3x2+ 4xy −4y2−14x+ 14y+ 13 = 0, we have A= 3,

B= 4, and C=−4. Therefore, the discriminant is:

B2−4AC = (4)2−4(3)(−4) = 16 + 48 = 64

Since the discriminant is positive and not equal to zero, the conic section is

a hyperbola.

Step 2: Find the equation of the axis The equation of the axis for a

hyperbola is given by 2y=mx +cwhere m=B

2Aand c=AF −

2BD

AC−

4B2.

Plugging in the values A= 3,B= 4,C=−4,D=−14,E= 14, and

F= 13, we get:

m=4

2(3) =2

c=(3)(13) −1

2(4)(−14)

(3)(−4) −1

4(4)2=39 + 28

−12 −4=67

−16 =−67

Therefore, the equation of the axis is:

2y=2

3x−67

Question 24

Question

Solve the following system of equations:

{x2+y2= 25

x−y= 3

Solution

Step 1: Solve the second equation for xin terms of y.

x−y= 3

x=y+ 3

Step 2: Substitute x=y+ 3 into the first equation.

(y+ 3)2+y2= 25

y2+ 6y+9+y2= 25

2y2+ 6y+ 9 −25 = 0

2y2+ 6y−16 = 0

y2+ 3y−8 = 0

Step 3: Find the roots of the quadratic equation y2+ 3y−8=0using the

quadratic formula.

y=−b±√b2−4ac

y=−3±√32−4·1·(−8)

2·1

y=−3±√9 + 32

y=−3±√41

Step 4: Calculate the corresponding values of x.

x=y+ 3

x1=−(3) + √41

2, x2=−(3) −√41

Therefore, the solutions to the system of equations are:

(−3 + √41

2,−3−√41

2)and (−3−√41

2,−3 + √41

Question 25

Question

Solve the following system of equations:

{4x2−9y2= 36

x−2y= 3

Solution

Step 1: Let’s solve the second equation for x:

x= 2y+ 3

Step 2: Substitute x= 2y+ 3 into the first equation:

4(2y+ 3)2−9y2= 36

Step 3: Expand and simplify the expression:

4(4y2+ 12y+ 9) −9y2= 36

16y2+ 48y+ 36 −9y2= 36

7y2+ 48y= 0

Step 4: Factor out y:

y(7y+ 48) = 0

Step 5: Set each factor to zero and solve for y:

y= 0 or 7y+ 48 = 0

Step 6: Solve for yin the second equation:

7y+ 48 = 0

7y=−48

y=−48

Step 7: Substitute y=−48

7back into x= 2y+ 3 to find x:

x= 2 (−48

7)+ 3

x=−96

7+21

x=−75

Step 8: The solution to the system of equations is:

{x=−75

y=−48

Step 6: Rearrange the equation to standard form by combining like terms:

h2−6h+k2+ 4k+ 10k+ 13 = 0

Step 7: Completing the square in terms of hand krespectively, we get:

(h−3)2+ (k+ 2)2= 0

Therefore, the standard form of the equation of the parabola is (y+ 2)2= 0 .

Question 2

Question

Solve the equation x2−y2+ 4x+ 6y−12 = 0 by completing the square, and

determine the type of conic section represented by the equation.

Solution

Step 1: Rearrange the equation by grouping the xand yterms together:

(x2+ 4x)−(y2−6y) = 12

Step 2: Complete the square for the xterms: 1. Take half of the coeﬀicient

of x, square it, and add/subtract it inside the parentheses:

(x2+ 4x+ 4) −(y2−6y) = 12 + 4

2. Simplify the square and the constant term:

(x+ 2)2−(y2−6y) = 16

Step 3: Complete the square for the yterms: 1. Take half of the coeﬀicient

of y, square it, and add/subtract it inside the parentheses:

(x+ 2)2−(y2−6y+ 9) = 16 + 9

2. Simplify the square and the constant term:

(x+ 2)2−(y−3)2= 25

Step 4: Compare the equation with the standard form of conic sections.

Since the signs of the squared terms are different, the equation represents a

hyperbola.

Question 3

Question

Let Cbe the conic section defined by the equation 4x2+ 9y2= 36. Find the

standard form of C.

Solution

Step 1: Divide both sides of the equation by 36 to get the standard form. Step

2: Rearrange the equation to isolate the terms.

Step 1:

4x2

36 +9y2

36 = 1

9+y2

4= 1

Step 2:

9+y2

4= 1

32−y2

22= 1

(x−0)2

32−(y−0)2

22= 1

Therefore, the standard form of the conic section Cis (x−0)2

32−(y−0)2

22= 1.

Question 4

Question

Solve the following system of equations:

{x2+y2= 25

x2−y2= 9

Solution

Step 1: We’ll start by solving the second equation in the system for x2in terms

of y.

x2= 9 + y2

Step 2: Substitute this expression for x2into the first equation of the system.

9 + y2+y2= 25

Step 3: Simplify the equation.

2y2= 16

Step 4: Solve for y.

y2= 8 =⇒y=±√8 = ±2√2

Step 5: Substitute the values of yback into x2= 9 + y2to find the corre-

sponding values of x. For y= 2√2,

x2= 9 + (2√2)2= 9 + 8 = 17 =⇒x=±√17

For y=−2√2,

x2= 9 + (−2√2)2= 9 + 8 = 17 =⇒x=±√17

Thus, the solutions to the system are:

(x, y) = (√17,2√2),(−√17,2√2),(√17,−2√2),(−√17,−2√2)

Question 5

Question

Find the standard form of the equation of the ellipse with foci at (−4,0) and

(4,0), and a major axis of length 8.

Solution

Step 1: The distance between the foci is equal to 2a, where ais the distance

from the center of the ellipse to a vertex along the major axis. In this case,

2a= 8 =⇒a= 4.

Step 2: The center of the ellipse is the midpoint of the line segment joining

the foci. The midpoint of the line segment with endpoints (−4,0) and (4,0) is

(0,0), so the center of the ellipse is at (0,0).

Step 3: The distance from the center of the ellipse to each focus is c. Since

cis 4 units from the center to each focus, then c= 4.

Step 4: The equation for an ellipse with center at the origin is x2

a2+y2

b2= 1,

where ais the distance from the center to a vertex along the major axis, and b

is the distance from the center to a vertex along the minor axis. Our equation

will be x2

42+y2

b2= 1.

Step 5: To find b, we use the relationship a2=b2+c2. Substituting the

values of aand c, we get 42=b2+42=⇒16 = b2+16 =⇒b2= 0 =⇒b= 0.

Step 6: Therefore, the equation for the ellipse is x2

16 +y2

0= 1, which simplifies

to x2

16 = 1 or x2= 16.

Step 7: The standard form of the equation of the ellipse is x2= 16.

Question 6

Question

Consider the conic section represented by the equation 4x2+y2−8x+6y+16 = 0.

Determine the type of conic section, find the center, vertices, and foci, and sketch

the graph of the conic section.

Solution

Step 1: Rewrite the given equation in standard form by completing the square

for both xand yterms.

4x2+y2−8x+ 6y+ 16 = 0

4(x2−2x) + y2+ 6y=−16

4(x2−2x+ 1) + y2+ 6y+ 9 = −16+4+9

4(x−1)2+ (y+ 3)2=−3

Step 2: Determine the type of conic section by examining the coeﬀicients of

the x2and y2terms. Since both coeﬀicients are positive and unequal, the conic

section is an ellipse.

Step 3: Find the center of the ellipse by looking at the completed square

form. The center of the ellipse is at (h, k) = (1,−3).

Step 4: Compute the vertices of the ellipse. The vertices are located aunits

to the left and right of the center, where ais the square root of the denominator

in the xterm. In this case, a=√3. Therefore, the vertices are at (1 −√3,−3)

and (1 + √3,−3).

Step 5: Calculate the distance from the center to the foci. The distance

is denoted by cand is found using the formula c=√a2−b2, where bis the

square root of the denominator in the yterm. Here, b=√3and a=√3. Thus,

c=√3.

Step 6: Locate the foci. The foci are situated cunits from the center along

the major axis. Hence, the foci are at (1 + √3,−3) and (1 −√3,−3).

Step 7: Sketch the ellipse by plotting the center, vertices, and foci on a

coordinate plane. Draw the ellipse passing through the vertices with the foci

located inside the ellipse.

Question 7

Question

Find the standard form of the equation of the parabola with focus F(0,3) and

directrix y=−3.

Solution

Step 1: First, we need to determine whether the parabola opens vertically or

horizontally. Since the directrix is parallel to the x-axis, the parabola opens

vertically.

Step 2: The distance between the vertex (h, k)of the parabola and the focus

is equal to the distance between the vertex and the directrix. Given that the

focus is F(0,3), and directrix is y=−3, the vertex has coordinates (0,0).

Step 3: Since the parabola opens vertically, the general form of the equation

of a parabola is (x−h)2= 4p(y−k). Since the vertex is (h, k) = (0,0), the

equation simplifies to x2= 4py.

Step 4: The distance between the vertex and the focus is the same as the

distance between the vertex and the directrix, which is |p|= 3. Therefore,

p= 3.

Step 5: Substitute the value of pinto the equation x2= 4py to obtain the

standard form of the equation. So, the equation of the parabola is x2= 12y.

Question 8

Question

Solve the following system of equations:

{x2+y2= 25

3x−4y= 6

Solution

Step 1: Let’s solve the second equation for xin terms of y:

3x−4y= 6

3x= 4y+ 6

x=4

3y+ 2

Step 2: Now substitute this expression for xinto the first equation:

3y+ 2)2

+y2= 25

9y2+16

3y+4+y2= 25

9y2+16

3y−21 = 0

Step 3: Multiply the equation by 9 to get rid of the fraction:

25y2+ 48y−189 = 0

Step 4: Now we have a quadratic equation. To solve for y, we can use the

quadratic formula:

y=−b±√b2−4ac

where a= 25,b= 48, and c=−189.

Step 5: Plugging in the values, we get:

y=−48 ±√482−4∗25 ∗(−189)

2∗25

y=−48 ±√2304 + 18900

y=−48 ±√21204

y=−48 ±2√5301

Step 6: This gives us two potential solutions for y. Now plug each value of

yback into the equation x=4

3y+ 2 to find the corresponding values of x.

Step 7: Finally, the solutions to the system of equations are the pairs of

(x, y)values that satisfy both equations simultaneously.

Question 9

Question

Let y=−2x2+ 8 be the equation of a parabola. Determine the coordinates of

the vertex, the focus, and the directrix of the parabola.

Solution

Step 1: We first rewrite the equation in standard form by completing the square

for x. Step 2: Rearrange the equation to center the parabola. Step 3: Compare

the new equation to the standard form of a parabola to identify the vertex,

the focus, and the directrix. Step 4: Write down the final answers for the

coordinates of the vertex, the focus, and the directrix.

Question 10

Question

A parabola has a vertex at the point (−2,3) and a focus at the point (4,3).

Determine the equation of the parabola in standard form.

Solution

Step 1: Recall that for a parabola with a horizontal axis of symmetry, the

equation in standard form is (x−h)2= 4p(y−k), where (h, k)is the vertex

and pis the distance from the vertex to the focus (or directrix).

Step 2: First, we identify the vertex as (h, k) = (−2,3).

Step 3: Next, we find the distance from the vertex to the focus. Since the

focus is at (4,3), the distance is p= 4 −(−2) = 6.

Step 4: Substitute the vertex and pinto the standard form equation to get

(x+ 2)2= 24(y−3).

Step 5: Simplify the equation to get the final answer in standard form:

(x+ 2)2= 24y−72.

Question 11

Question

Solve the equation 3x2−2y2+ 18x+ 4y−22 = 0 for y.

Solution

Step 1: Rewrite the given equation in standard form by completing the square

for both xand yterms.

3x2−2y2+ 18x+ 4y−22 = 0

3(x2+ 6x)−2(y2−2y) = 22

3(x2+ 6x+ 9) −2(y2−2y+ 1) = 22 + 3(9) −2(1)

3(x+ 3)2−2(y−1)2= 49

Step 2: Divide both sides of the equation by 49 to obtain 1on the right side.

3(x+ 3)2

49 −2(y−1)2

49 = 1

(x+ 3)2

(7)2−(y−1)2

(7/√2)2= 1

Step 3: Compare the equation obtained with the standard form of hyperbola

to find values of aand b. The equation of a hyperbola in standard form is

(x−h)2

a2−(y−k)2

b2= 1. Comparing it with the given equation, we have a= 7 and

b=7

√2.

Step 4: The standard form of a hyperbola centered at (h, k)with major axis

along the x-axis is (x−h)2

a2−(y−k)2

b2= 1. Thus, the equation of hyperbola in

standard form for this problem is (x+3)2

(7)2−(y−1)2

(7/√2)2= 1.

Question 12

Question

Solve the following system of equations:

{x2+ 4y2= 5

x2−y2= 1

Solution

Step 1: Start by isolating x2in the second equation:

x2=y2+ 1

Step 2: Substitute x2in terms of y2from the second equation into the first

equation:

(y2+1)+4y2= 5

Step 3: Simplify the equation by combining like terms:

5y2+ 1 = 5

Step 4: Subtract 1 from both sides of the equation:

5y2= 4

Step 5: Divide by 5 to solve for y2:

y2=4

Step 6: Take the square root of both sides to solve for y:

y=±√4

Step 7: Simplify the square root to obtain the values of y:

y=±2

√5

Step 8: Substitute the values of yback into the equation x2=y2+1 to solve

for x:

x2=(2

√5)2

+ 1 or x2=(−2

√5)2

+ 1

Step 9: Simplify to find the values of x:

x=±√1

5+ 1 or x=±√1

5+ 1

Step 10: Simplify further to get the final solutions for x:

x=±√6

5or x=±√6

Therefore, the solutions to the system of equations are:

(x, y) = (√6

5,2

√5),(−√6

5,−2

√5),(√6

5,−2

√5),(−√6

5,2

√5)

Question 13

Question

Solve the system of equations for xand y:

{3x2−4xy +y2= 10

2x2+ 3xy −5y2=−7

Solution

Step 1: Multiply the first equation by 5 and the second equation by 2 to make

the coeﬀicient of y2the same in both equations:

{15x2−20xy + 5y2= 50

4x2+ 6xy −10y2=−14

Step 2: Add the two equations together to eliminate the term y2:

19x2−14y2= 36

Step 3: This equation represents a hyperbola. To simplify, divide by 36:

(√36

19 )2−y2

(√36

14 )2= 1

6√19

19 −y2

6√14

= 1

6√19

19 −y2

3√14

= 1

Step 4: Comparing this equation with the standard form of a hyperbola,

a2−y2

b2= 1, we see that a=6√19

19 and b=3√14

Therefore, the solution to the system of equations is a hyperbola with a=

6√19

19 and b=3√14

Question 14

Question

Find the standard form of the equation of the hyperbola that satisfies the given

conditions: foci at (±5,0) and passes through the point (4,3).

Solution

Step 1: Determine the center of the hyperbola. Since the foci are on the x-axis,

the center of the hyperbola will also lie on the x-axis. The x-coordinate of the

center is the average of the x-coordinates of the foci:

Center =(−5+5

2,0)= (0,0)

Step 2: Write the equation of the hyperbola in standard form. The general

form of the equation of a hyperbola centered at (h, k)is (x−h)2

a2−(y−k)2

b2= 1.

We need to find the values of aand b.

Step 3: Determine the distance from the center to one of the foci, which is

the value of c. In this case, c= 5.

Step 4: Use the distance formula to determine the equation of the ellipse.

The distance between the point (4,3) and a focus (5,0) is equal to 2a(the

distance from the center to a vertex). Therefore,

√(4 −5)2+ (3 −0)2= 2a

√(−1)2+ 32= 2a

√10 = 2a

a=√10

Step 5: Find the value of busing a,c, and the relationship between a,b, and

cin hyperbolas:

c2=a2+b2

52=(√10

2)2

+b2

25 = 10

4+b2

b2= 25 −10

b2=90

b=3√10

Step 6: Write the equation of the hyperbola in standard form.

(√10/2)2−y2

(3√10/2)2= 1

5/2−y2

45/2= 1

2x2−2y2

9= 1

Question 15

Question

Solve the following system of equations:

{x2+y2= 25

x−y= 3

Solution

Step 1: We start by solving the second equation for xin terms of y.

x−y= 3

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+ 9 = 25

2y2+ 6y−16 = 0

Step 4: Solve the quadratic equation 2y2+ 6y−16 = 0 using the quadratic

formula.

y=−b±√b2−4ac

Step 5: Plug in a= 2,b= 6, and c=−16 into the quadratic formula.

y=−6±√62−4(2)(−16)

2(2)

y=−6±√36 + 128

y=−6±√164

Step 6: Simplify the expression under the square root.

√164 = √4×41 = 2√41

Step 7: Substitute this back into the equation for y.

y=−6±2√41

y=−3±√41

Step 8: Solve for xusing x=y+ 3.

x=−3±√41

2+ 3

x=−3±√41

2+6

x=3±√41

Step 9: So, the solutions to the system of equations are:

{x=3+√41

2, y =−3+√41

x=3−√41

2, y =−3−√41

Question 16

Question

Determine the standard form of the equation of the parabola with focus at

F(3,4) and directrix x= 1.

Solution

Step 1: Recall the definition of a parabola: The set of all points that are

equidistant from a fixed point (the focus) and a fixed line (the directrix).

Step 2: Determine the vertex of the parabola. Since the focus is at F(3,4)

and the directrix is x= 1, the vertex lies halfway between the focus and the

directrix. Thus, the vertex is at V(Fx+D

2,Fy

2), where Dis the x-coordinate of

the directrix. Therefore, the vertex is V(3+1

2,4

2)= (2,2).

Step 3: Determine the distance between the focus and the vertex. This

distance is called the focal length, denoted by |p|. Since the focus is at F(3,4)

and the vertex is at (2,2),|p|=|Fx−Vx|=|3−2|= 1.

Step 4: Determine the equation of the parabola. Since the directrix is vertical

and to the left of the vertex, the parabola opens to the right. The standard form

of the equation of the parabola in this case is (x−h)2= 4p(y−k), where (h, k)

is the vertex.

Step 5: Substitute the known values into the standard form equation: (x−

2)2= 4(1)(y−2).

Step 6: Simplify the equation: (x−2)2= 4(y−2).

Therefore, the standard form of the equation of the parabola is (x−2)2=

4(y−2).

Question 17

Question

Solve the equation for y:4x2+ 9y2+ 8x−36y+ 23 = 0.

Solution

Step 1: Rearrange the equation by grouping the xterms and completing the

square for x.

4x2+ 9y2+ 8x−36y+ 23 = 0

4x2+ 8x+ 9y2−36y=−23

4(x2+ 2x) + 9(y2−4y) = −23

4(x2+ 2x+ 1) + 9(y2−4y+ 4) = −23 + 4(1) + 9(4)

4(x+ 1)2+ 9(y−2)2= 10

Step 2: Divide by 10 to get equation in standard form.

4(x+ 1)2

10 +9(y−2)2

10 = 1

(x+ 1)2

+(y−2)2

= 1

Step 3: Identify the ellipse’s center, vertices, foci, and eccentricity. The

equation (x−h)2

a2+(y−k)2

b2= 1 represents an ellipse centered at (h, k)with semi-

major axis aand semi-minor axis b. Comparing with our equation, we find:

(h, k) = (−1,2)

a=√10

4=√10

b=√10

9=√10

The center of the ellipse is (−1,2), semi-major axis is √10

2, and semi-minor

axis is √10

3. Hence, the ellipse’s vertices are (−1±√10

2,2), foci are (−1±√6,2),

and eccentricity is e=√1−b2

a2=√1−10

4·9=√14

36 =√7

Question 18

Question

Solve the equation 3x2+ 4y2+ 6x−16y+ 9 = 0 by completing the square.

Identify the conic section represented by the equation.

Solution

Step 1: Rewrite the equation by grouping xterms together and yterms together:

(3x2+ 6x) + (4y2−16y) + 9 = 0

Step 2: Complete the square for the xand yterms separately. For the x

terms:

(3x2+ 6x)

3(x2+ 2x)

3(x2+ 2x+ 1 −1)

3[(x+ 1)2−1]

3(x+ 1)2−3

For the yterms:

(4y2−16y)

4(y2−4y)

4(y2−4y+ 4 −4)

4[(y−2)2−4]

4(y−2)2−16

Adding the completed square expressions together, we have:

3(x+ 1)2−3 + 4(y−2)2−16 + 9 = 0

Step 3: Combine like terms:

3(x+ 1)2+ 4(y−2)2−10 = 0

Step 4: Divide the entire equation by 10 to simplify:

10(x+ 1)2+4

10(y−2)2−1 = 0

Step 5: Identify the conic section represented by the equation. Since the

coeﬀicients of (x+1)2and (y−2)2have the same sign, and the larger coeﬀicient

is on the xterm, the conic section is an ellipse.

Question 19

Question

Solve the system of equations:

{2x2+y2= 16

x2−2y2= 3

Solution

Step 1: Let’s first rewrite the system of equations in standard form by expanding

the terms: {2x2+y2= 16 (1)

x2−2y2= 3 (2)

Step 2: To eliminate a variable, let’s multiply equation (2) by 2:

2(x2−2y2) = 2(3)

2x2−4y2= 6

Step 3: Subtract equation (1) from the new equation we obtained:

(2x2−4y2)−(2x2+y2) = 6 −16

−5y2=−10

Step 4: Solve for y2:

y2= 2

Step 5: Substitute y2= 2 back into equation (1):

2x2+ 2 = 16

2x2= 14

x2= 7

Step 6: Take the square root of both sides to solve for x:

x=±√7

Therefore, the solution to the system of equations is:

(x, y) = (√7,√2),(−√7,√2)

Question 20

Question

Solve the system of equations:

{x2+y2−4x+ 6y+ 4 = 0

3x−4y+ 2 = 0

Solution

Step 1: We can rewrite the first equation in the standard form of a circle by

completing the square for xand yterms:

x2+y2−4x+ 6y+ 4 = 0

(x2−4x)+(y2+ 6y) = −4

(x2−4x+4)+(y2+ 6y+ 9) = −4+4+9

(x−2)2+ (y+ 3)2= 9

Step 2: The equation (x−2)2+ (y+ 3)2= 9 represents a circle with center

at (2,−3) and radius 3.

Step 3: Now, we substitute 3x−4y+ 2 = 0 into the equation of the circle:

3x−4y+ 2 = 0

3x= 4y−2

x=4

3y−2

Substituting xback into the equation of the circle, we get:

3y−2

3−2)2+ (y+ 3)2= 9

Step 4: Solving for yin the equation above may involve some algebraic

manipulations to simplify the expression and solve for y.

y=−2or y=−4

Step 5: Finally, substituting the values of yback into x=4

3y−2

3, we find

the corresponding values of xfor each yvalue.

Therefore, the solutions to the system of equations are:

(x, y) = (2

3,−2)and (x, y) = (2

3,−4)

Question 21

Question

Find the equation of the hyperbola with foci at (−2,1) and (4,1) and vertices

at (−5,1) and (7,1).