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College Algebra : Graphs

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Example Questions

Example Question #5 : Ellipses

Give the eccentricity of the ellipse of the equation

$\frac{x^{2}}{25}+ \frac{y^{2}}{169} = 1$

Possible Answers:

$\frac{5}{12}$

$\frac{12}{13}$

$\frac{144}{169}$

$\frac{5}{13}$

$\frac{25}{144}$

Correct answer:

$\frac{12}{13}$

Explanation:

This ellipse is in standard form

$\frac{x^{2}}{a^{2}}+ \frac{y^{2}}{b^{2}} = 1$

where $b^{2} > a^{2}$ . This is a vertical ellipse, whose foci are

$c= \sqrt{b^{2}-a^{2}}$

units from its center in a vertical direction.

The eccentricity of this ellipse can be calculated by taking the ratio $\frac{c}{b}$ , or, equivalently, $\frac{\sqrt{b^{2}-a^{2}}}{b}$ . Set $b^{2} = 169$ - making b= 13 - and $a^{2} = 25$ . The eccentricity is calculated to be

$\frac{\sqrt{169-25}}{13} =\frac{ \sqrt{144}}{13} = \frac{12}{13}$ .

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Example Question #11 : Ellipses

The graph of the equation

$16x^{2}+12y^{2}-96x+48y+384=0$

is an example of which conic section?

Possible Answers:

A vertical ellipse

The equation has no graph.

A vertical hyperbola

A horizontal ellipse

A horizontal hyperbola

Correct answer:

The equation has no graph.

Explanation:

The quadratic coefficients in this general form of a conic equation are 16 and 12. They are of the same sign, making its graph, if it exists, an ellipse.

To determine whether this ellipse is horizontal or vertical, rewrite this equation in standard form

$\frac{\left ( x- h\right )^{2}}{a^{2}} + \frac{\left ( y- k\right )^{2}}{b^{2}} = 1$

as follows:

Subtract 384 from both sides:

$16x^{2}+12y^{2}-96x+48y+384=0$

$16x^{2}+12y^{2}-96x+48y= -384$

Separate the and terms and group them:

$16x^{2} + 12y^{2} -96x-48y =-384$

$(16x^{2} -96x) + (12y^{2}-48y) =-384$

Distribute out the quadratic coefficients:

$16 ( x^{2} -6x ) +12 ( y^{2}-4y) =-384$

Complete the square within each quadratic expression by dividing each linear coefficient by 2 and squaring the quotient.

Since $\left (\frac{-6}{2} \right )^{2}= 9$ and $\left (\frac{-4}{2} \right )^{2}= 4$ , we get

$16 ( x^{2} -6x \textbf{+ 9} ) +12 ( y^{2}-4y\textbf{ + 4}) =-384+ \textbf{?}+ \textbf{?}$

Balance this equation, adjusting for the distributed coefficients:

$16 ( x^{2} -6x + 9 ) +12 ( y^{2}-4y + 4 ) =-384+ 16(9)+ 12(4)$

$16 ( x^{2} -6x + 9 ) +12 ( y^{2}-4y + 4 ) =-192$

The perfect square trinomials are squares of binomials, by design; rewrite them as such:

$16 ( x-3 )^{2} +12 ( y -2 ) ^{2} =-192$

Divide by :

$\frac{16 ( x-3 )^{2}}{-192} +\frac{12 ( y -2 ) ^{2} }{-192}=\frac{-192}{-192}$

$\frac{( x-3 )^{2}}{-12} +\frac{( y -2 ) ^{2} }{-16}=1$

Recall that the standard form of an ellipse is

$\frac{\left ( x- h\right )^{2}}{a^{2}} + \frac{\left ( y- k\right )^{2}}{b^{2}} = 1$

This requires both denominators to be positive. In the standard form of the given equation, they are not. Therefore, the equation has no real ordered pairs as solutions, and it does not have a graph on the coordinate plane.

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Example Question #121 : College Algebra

Give the foci of the ellipse of the equation

$\frac{(x- 2)^{2}}{64} + \frac{(y-6)^{2}}{49} = 1$ .

Round your coordinates to the nearest tenth, if applicable.

Possible Answers:

(2, -4.6),(2, 16.6)

(2, 2.1), (2, 9.9)

(-1.9, 6), (5.9,6 )

(-8.6, 6), (12.6, 6)

None of the other choices gives the correct response.

Correct answer:

(-1.9, 6), (5.9,6 )

Explanation:

$\frac{(x- h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1$

is the standard form of an ellipse with center (h,k) . Also, since in the given equation, $a^{2} = 64$ and $b^{2} = 49$ - that is, $a^{2}> b^{2}$ , the ellipse is horizontal.

The foci of a horizontal ellipse are located at

$(h \pm c ,k)$ ,

where

$c= \sqrt{a^{2}-b^{2}} = \sqrt{64-49} = \sqrt{15} \approx 3.9$

Setting $h = 2, k = 6, c \approx 3.9$ , the foci are at

$(2 \pm 3.9, 6)$ , or

(-1.9, 6) and (5.9,6 ) .

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Example Question #4 : Conic Sections

Using the information below, determine the equation of the hyperbola.

Foci: (-5,4) and (7,4)

Eccentricity: $\frac{3}{2}$

Possible Answers:

$\small \small \frac{(x-1)^{2}}{16} - \frac{(y-4)^{2}}{20}=1$

$\small \small \small \frac{(x-1)^{2}}{20} - \frac{(y-4)^{2}}{16}=1$

$\small \small \small \small \frac{(x-1)^{2}}{16} - \frac{(y-4)^{2}}{36}=1$

$\small \small \small \frac{(x-4)^{2}}{16} - \frac{(y-1)^{2}}{20}=1$

$\small \small \small \small \small \frac{(x-4)^{2}}{20} - \frac{(y-1)^{2}}{36}=1$

Correct answer:

$\small \small \frac{(x-1)^{2}}{16} - \frac{(y-4)^{2}}{20}=1$

Explanation:

General Information for Hyperbola:

Equation for horizontal transverse hyperbola:

$\small \frac{(x-h)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}}=1$

Distance between foci =

Distance between vertices =

Eccentricity = c/a

$\small a^{2}+b^{2}=c^{2}$

Center: (h, k)

First determine the value of c. Since we know the distance between the two foci is 12, we can set that equal to .

12=2c

$\small c= 6$

Next, use the eccentricity equation and the value of the eccentricity provided in the question to determine the value of a.

Eccentricity = $\small \small c/a=3/2$

$\small \frac{c}{a}=\frac{3}{2}=\frac{6}{a}$

$\small 12=3a$

$\small a=4$

$\small \small a^{2}=16$

Determine the value of $\small b^{2}$

$\small a^{2}+b^{2}=c^{2}$

$\small \small 4^{2}+b^{2}=6^{2}$

$\small \small b^{2}=20$

Determine the center point to identify the values of h and k. Since the y coordinate of the foci are 4, the center point will be on the same line. Hence, $\small k = 4$ .

Since center point is equal distance from both foci, and we know that the distance between the foci is 12, we can conclude that $\small h=1$

Center point: $\small \small (h, k)= (1,4)$

Thus, the equation of the hyperbola is:

$\small \small \frac{(x-1)^{2}}{16} - \frac{(y-4)^{2}}{20}=1$

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Example Question #1 : Conic Sections

Using the information below, determine the equation of the hyperbola.

Foci: (1, 8) and (9, 8)

Eccentricity:

Possible Answers:

$\small \small \small \small \small \small \small \frac{(x-5)^{2}}{4} - \frac{(y-8)^{2}}{16}=1$

$\small \small \small \small \small \small \small \small \frac{(x-8)^{2}}{12} - \frac{(y-5)^{2}}{16}=1$

$\small \small \small \small \small \small \small \frac{(x-5)^{2}}{12} - \frac{(y-8)^{2}}{4}=1$

$\small \small \small \small \small \small \frac{(x-5)^{2}}{4} - \frac{(y-8)^{2}}{12}=1$

$\small \small \small \small \small \small \small \frac{(x-8)^{2}}{4} - \frac{(y-5)^{2}}{12}=1$

Correct answer:

$\small \small \small \small \small \small \frac{(x-5)^{2}}{4} - \frac{(y-8)^{2}}{12}=1$

Explanation:

General Information for Hyperbola:

Equation for horizontal transverse hyperbola:

$\small \frac{(x-h)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}}=1$

Distance between foci =

Distance between vertices =

Eccentricity = c/a

$\small a^{2}+b^{2}=c^{2}$

Center: (h, k)

First determine the value of c. Since we know the distance between the two foci is 8, we can set that equal to .

$\small 8=2c$

$\small \small c= 4$

Next, use the eccentricity equation and the value of the eccentricity provided in the question to determine the value of a.

Eccentricity = $\small \small \small \small c/a=2$

$\small \small \frac{c}{a}=\frac{2}{1}=\frac{4}{a}$

$\small \small 4=2a$

$\small \small a=2$

$\small \small \small a^{2}=4$

Determine the value of $\small b^{2}$

$\small a^{2}+b^{2}=c^{2}$

$\small \small \small 2^{2}+b^{2}=4^{2}$

$\small \small \small b^{2}=12$

Determine the center point to identify the values of h and k. Since the y coordinate of the foci are 8, the center point will be on the same line. Hence, $\small \small k = 8$ .

Since center point is equal distance from both foci, and we know that the distance between the foci is 8, we can conclude that $\small \small h=5$

Center point: $\small \small \small (h, k)= (5, 8 )$

Thus, the equation of the hyperbola is:

$\small \small \small \small \small \small \frac{(x-5)^{2}}{4} - \frac{(y-8)^{2}}{12}=1$

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Example Question #41 : Graphs

Find the foci of the hyperbola with the following equation:

$\frac{x^2}{27}-\frac{y^2}{9}=1$

Possible Answers:

(0, 6)(0, -6)

(0, 9)(0, -9)

(6, 0)(-6, 0)

(12, 0)(-12, 0)

Correct answer:

(6, 0)(-6, 0)

Explanation:

Recall that the standard formula of a hyperbola can come in two forms:

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ and

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}-=1$ , where the centers of both hyperbolas are (h, k) .

When the term with is first, that means the foci will lie on a horizontal transverse axis.

When the term with is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

$c=\sqrt{a^2+b^2}$

For a hyperbola with a horizontal transverse access, the foci will be located at (h+c, k) and (h-c, k) .

For a hyperbola with a vertical transverse access, the foci will be located at (h, k+c) and (h, k-c) .

For the given hypebola in the question, the transverse axis is horizontal and its center is located at (0, 0) .

Next, find .

$c=\sqrt{27+9}=\sqrt{36}=6$

The foci are then located at (6, 0) and (-6, 0) .

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Example Question #201 : Conic Sections

Find the foci of a hyperbola with the following equation:

$\frac{x^2}{36}-\frac{y^2}{64}=1$

Possible Answers:

(6, 0)(-6, 0)

(0, 8)(0, -8)

(10, 0)(-10, 0)

(0, 10)(0, -10)

Correct answer:

(10, 0)(-10, 0)

Explanation:

Recall that the standard formula of a hyperbola can come in two forms:

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ and

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}-=1$ , where the centers of both hyperbolas are (h, k) .

When the term with is first, that means the foci will lie on a horizontal transverse axis.

When the term with is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

$c=\sqrt{a^2+b^2}$

For a hyperbola with a horizontal transverse access, the foci will be located at (h+c, k) and (h-c, k) .

For a hyperbola with a vertical transverse access, the foci will be located at (h, k+c) and (h, k-c) .

For the given hypebola in the question, the transverse axis is horizontal and its center is located at (0, 0) .

Next, find .

$c=\sqrt{64+36}=\sqrt{100}=10$

The foci are then located at (10, 0) and (-10, 0) .

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Example Question #42 : Graphs

Find the foci of the hyperbola with the following equation:

$\frac{(x+2)^2}{16}-\frac{(y-2)^2}{9}=1$

Possible Answers:

(-2, 2)(7, 2)

(3, 2)(-7, 2)

(-2, -7)(-2, -3)

(-2, 7)(-2, -3)

Correct answer:

(3, 2)(-7, 2)

Explanation:

Recall that the standard formula of a hyperbola can come in two forms:

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ and

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}-=1$ , where the centers of both hyperbolas are (h, k) .

When the term with is first, that means the foci will lie on a horizontal transverse axis.

When the term with is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

$c=\sqrt{a^2+b^2}$

For a hyperbola with a horizontal transverse access, the foci will be located at (h+c, k) and (h-c, k) .

For a hyperbola with a vertical transverse access, the foci will be located at (h, k+c) and (h, k-c) .

For the given hypebola in the question, the transverse axis is horizontal and its center is located at (-2,2) .

Next, find .

$c=\sqrt{16+9}=\sqrt{25}=5$

The foci are then located at (3, 2) and (-7, 2) .

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Example Question #43 : Graphs

Find the foci of the hyperbola with the following equation:

50x^2-14y^2-400x+56y+44=0

Possible Answers:

(14, 2)(4, 2)

(10, 2)(-2, 2)

(4, 10)(4, -6)

(12, 2)(-4, 2)

Correct answer:

(12, 2)(-4, 2)

Explanation:

Recall that the standard formula of a hyperbola can come in two forms:

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ and

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}-=1$ , where the centers of both hyperbolas are (h, k) .

First, put the given equation in the standard form of the equation of a hyperbola.

Group the terms together and the terms together.

50x^2-14y^2-400x+56y+44=0

50x^2-400x-14y^2+56y+44=0

Next, factor out from the terms and from the terms.

50(x^2-8x)-14(y^2-4y)+44=0

From here, complete the squares. Remember to add the same amount to both sides of the equation!

50(x^2-8x+16)-14(y^2-4y+4)+44=744

Subtract from both sides.

50(x^2-8x+16)-14(y^2-4y+4)=700

Divide both sides by 700 .

$\frac{x^2-8x+16}{14}-\frac{y^2-4y+4}{50}=1$

Factor both terms to get the standard form for the equation of a hyperbola.

$\frac{(x-4)^2}{14}-\frac{(y-2)^2}{50}=1$

When the term with is first, that means the foci will lie on a horizontal transverse axis.

When the term with is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

$c=\sqrt{a^2+b^2}$

For a hyperbola with a horizontal transverse access, the foci will be located at (h+c, k) and (h-c, k) .

For a hyperbola with a vertical transverse access, the foci will be located at (h, k+c) and (h, k-c) .

For the given hypebola in the question, the transverse axis is horizontal and its center is located at (4,2) .

Next, find .

$c=\sqrt{14+50}=\sqrt{64}=8$

The foci are then located at (12, 2) and (-4, 2) .

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Example Question #203 : Conic Sections

Find the foci of the hyperbola with the following equation:

12x^2-4y^2-216x-8y+920=0

Possible Answers:

(13, -1)(7, -1)

(13, -1)(5, -1)

(9, 4)(9, -6)

(9, -1)(9, -7)

Correct answer:

(13, -1)(5, -1)

Explanation:

Recall that the standard formula of a hyperbola can come in two forms:

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ and

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}-=1$ , where the centers of both hyperbolas are (h, k) .

First, put the given equation in the standard form of the equation of a hyperbola.

Group the terms together and the terms together.

12x^2-4y^2-216x-8y+920=0

12x^2-216x-4y^2-8y+920=0

Next, factor out from the terms and from the terms.

12(x^2-18x)-4(y^2+2y)+920=0

From here, complete the squares. Remember to add the same amount to both sides of the equation!

12(x^2-18x+81)-4(y^2+2y+1)+920=968

Subtract 920 from both sides.

12(x^2-18x+81)-4(y^2+2y+1)=48

Divide both sides by .

$\frac{x^2-18x+81}{4}-\frac{y^2+2y+1}{12}=1$

Factor both terms to get the standard form for the equation of a hyperbola.

$\frac{(x-9)^2}{4}-\frac{(y+1)^2}{12}=1$

When the term with is first, that means the foci will lie on a horizontal transverse axis.

When the term with is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

$c=\sqrt{a^2+b^2}$

For a hyperbola with a horizontal transverse access, the foci will be located at (h+c, k) and (h-c, k) .

For a hyperbola with a vertical transverse access, the foci will be located at (h, k+c) and (h, k-c) .

For the given hypebola in the question, the transverse axis is horizontal and its center is located at (9, -1) .

Next, find .

$c=\sqrt{12+4}=\sqrt{16}=4$

The foci are then located at (13, -1) and (5, -1) .

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College Algebra : Graphs

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #5 : Ellipses

Example Question #11 : Ellipses

Example Question #121 : College Algebra

Example Question #4 : Conic Sections

Example Question #1 : Conic Sections

Example Question #41 : Graphs

Example Question #201 : Conic Sections

Example Question #42 : Graphs

Example Question #43 : Graphs

Example Question #203 : Conic Sections

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