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Precalculus : Hyperbolas and Ellipses

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Example Questions

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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Example Question #111 : Understand Features Of Hyperbolas And Ellipses

Find the foci of the hyperbola with the following equation:

20y^2-16x^2-120y+64x-204=0

Possible Answers:

(8, 3)(-4, 3)

(2, 9)(2, -3)

(2, 9)(9, 2)

(2, 9)(2, 3)

Correct answer:

(2, 9)(2, -3)

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.

20y^2-16x^2-120y+64x-204=0

20y^2-120y-16x^2+64x-204=0

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

20(y^2-6y)-4(x^2-4x)-204=0

Complete the squares. Remember to add the same amount to both sides of the equation!

20(y^2-6y+9)-4(x^2-4x+4)-204=116

Add 204 to both sides.

20(y^2-6y+9)-4(x^2-4x+4)=320

Divide both sides by 320 .

$\frac{y^2-6y+9}{16}-\frac{x^2-4x+4}{20}=1$

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

$\frac{(y-3)^2}{16}-\frac{(x-2)^2}{20}=1$

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 vertical and its center is located at (2,3) .

Next, find .

$c=\sqrt{16+20}=\sqrt{36}=6$

The foci are then located at (2, 9) and (2, -3) .

Report an Error

Example Question #112 : Understand Features Of Hyperbolas And Ellipses

Find the foci of the hyperbola with the following equation:

22y^2-27x^2+220y-54x-71=0

Possible Answers:

(6, -5)(-8, -5)

(-12, 12)(-1, 12)

(-1, 2)(1, 2)

(-1, 2)(-1, -12)

Correct answer:

(-1, 2)(-1, -12)

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.

22y^2-27x^2+220y-54x-71=0

22y^2+220y-27x^2-54x-71=0

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

22(y^2+10y)-27(x^2+2x)-71=0

Now we can complete the squares. Remember to add the same amount to both sides of the equation!

22(y^2+10y+25)-27(x^2+2x+1)-71=523

Add to both sides.

22(y^2+10y+25)-27(x^2+2x+1)-71=594

Divide both sides by 594 .

$\frac{y^2+10y+25}{27}-\frac{x^2+2x+1}{22}=1$

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

$\frac{(y+5)^2}{27}-\frac{(x+1)^2}{22}=1$

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 vertical and its center is located at (-1,-5) .

Next, find .

$c=\sqrt{22+27}=\sqrt{49}=7$

The foci are then located at (-1, 2) and (-1, -12) .

Report an Error

Example Question #1781 : Pre Calculus

Find the foci of the hyperbola with the following equation:

$\frac{(y-7)^2}{9}-\frac{(x+6)^2}{7}=1$

Possible Answers:

(-2, 7)(-10, 7)

(11, -6)(7, -6)

(11, -6)(3, -6)

(-6, 11)(-6, 3)

Correct answer:

(-6, 11)(-6, 3)

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 vertical and its center is located at (-6,7) .

Next, find .

$c=\sqrt{9+7}=\sqrt{16}=4$

The foci are then located at (-6,11) and (-6,3) .

Report an Error

Example Question #111 : Understand Features Of Hyperbolas And Ellipses

Find the foci of the hyperbola with the following equation:

$\frac{(y-1)^2}{14}-\frac{(x-5)^2}{22}=1$

Possible Answers:

(5, 7)(5, -5)

(11, 5)(-1, 5)

(11, 1)(-1, 1)

(-6, 6)(-6, 11)

Correct answer:

(5, 7)(5, -5)

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 vertical and its center is located at (5,1) .

Next, find .

$c=\sqrt{14+22}=\sqrt{36}=6$

The foci are then located at (5, 7) and (5, -5) .

Report an Error

Example Question #113 : Hyperbolas And Ellipses

Find the foci of the hyperbola with the following equation:

$\frac{y^2}{14}-\frac{x^2}{2}=1$

Possible Answers:

(4, 0)(-4, 0)

$(0, \sqrt{14})(0, -\sqrt{14})$

(0, 4)(0, -4)

$(\sqrt2, 0)(-\sqrt2, 0)$

Correct answer:

(0, 4)(0, -4)

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 vertical and its center is located at (0,0) .

Next, find .

$c=\sqrt{14+2}=\sqrt{16}=4$

The foci are then located at (0, 4) and (0, -4) .

Report an Error

Example Question #1793 : Pre Calculus

Find the foci for the hyperbola with the following equation:

$\frac{y^2}{12}-\frac{x^2}{13}=1$

Possible Answers:

$(2\sqrt3, 0)(-2\sqrt3, 0)$

(0, 5)(0, -5)

(5, 0)(-5, 0)

$(0, \sqrt{13})(0, -\sqrt{13})$

Correct answer:

(0, 5)(0, -5)

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 vertical and its center is located at (0,0) .

Next, find .

$c=\sqrt{12+13}=\sqrt{25}=5$

The foci are then located at (0, 5) and (0, -5) .

Report an Error

Example Question #112 : Understand Features Of Hyperbolas And Ellipses

Which point is one of the foci of the hyperbola $\frac{(x-5)^2 }{ 9 } - \frac{(y+1)^2 }{ 2 } = 1$ ?

Possible Answers:

$(5 + \sqrt{11}, -1 )$

$(5, 1 - \sqrt{11})$

$(\sqrt{11} - 5, 1 )$

$(5 + \sqrt{7} , 1 )$

$(5, -1 + \sqrt{7})$

Correct answer:

$(5 + \sqrt{11}, -1 )$

Explanation:

To find the foci of a hyperbola, first determine a and b, and then use the relationship b^2 = c^2 - a^2

In this case, the major axis is horizontal since x comes first, so a^2 = 9 and b^2 = 2 .

Solve for c: 2 = c^2 - 9 add 9 to both sides

11 = c^2 take the square root

$\sqrt{11} = c$

Since the center is (5, -1) and the major axis is the horizontal one, our foci are $(5 \pm \sqrt{11} , -1)$ . The only choice that works is

$(5 + \sqrt{11}, -1 )$ .

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Precalculus : Hyperbolas and Ellipses

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #42 : Graphs

Example Question #43 : Graphs

Example Question #203 : Conic Sections

Example Question #111 : Understand Features Of Hyperbolas And Ellipses

Example Question #112 : Understand Features Of Hyperbolas And Ellipses

Example Question #1781 : Pre Calculus

Example Question #111 : Understand Features Of Hyperbolas And Ellipses

Example Question #113 : Hyperbolas And Ellipses

Example Question #1793 : Pre Calculus

Example Question #112 : Understand Features Of Hyperbolas And Ellipses

All Precalculus Resources

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