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Precalculus : Hyperbolas and Ellipses

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Example Questions

Example Question #1775 : Pre Calculus

Find the eccentricty of the following hyperbola:

13x^2-12y^2-52x+24y-116=0

Possible Answers:

$\frac{2\sqrt3}{3}$

$\frac{12}{13}$

$\frac{5}{12}$

$\frac{5\sqrt{12}}{12}$

Correct answer:

$\frac{5\sqrt{12}}{12}$

Explanation:

Recall the standard form of the equation of a horizontal hyperbola:

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}$ , where (h,k) is the center

Start by putting the given equation into the standard form of the equation of a horizontal hyperbola.

Group the terms and terms together.

13x^2-12y^2-52x+24y-116=0

13x^2-52x-12y^2+24y-116=0

Factor out from the terms and from the terms.

13(x^2-4x)-12(y^2-2y)=116

Complete the squares. Remember to add the same number on both sides!

13(x^2-4x+4)-12(y^2-2y+1)=156

Divide both sides by 156 .

$\frac{x^2-4x+4}{12}-\frac{y^2-2y+1}{13}=1$

Factor both terms to get the standard equation.

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

For a horizontal hyperbola, use the following equation to find the eccentricity:

$e=\frac{c}{a}$ , where $c=\sqrt{a^b+b^2}$

For the given hyperbola,

$a=\sqrt{12}$ and $b=\sqrt{13}$

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

Thus,

$e=\frac{5}{\sqrt{12}}=\frac{5\sqrt{12}}{12}$

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Example Question #101 : Understand Features Of Hyperbolas And Ellipses

Find the eccentricity of the following hyperbola:

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

Possible Answers:

$\frac{3}{2}$

$\frac{\sqrt2}{13}$

$\frac{\sqrt{13}}{3}$

$\frac{\sqrt{13}}{2}$

Correct answer:

$\frac{\sqrt{13}}{2}$

Explanation:

Recall the standard form of the equation of a vertical hyperbola:

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$ , where (h,k) is the center

For a horizontal hyperbola, use the following equation to find the eccentricity:

$e=\frac{c}{b}$ , where $c=\sqrt{a^b+b^2}$

For the given hyperbola,

a=3 and b=2

$c=\sqrt{9+4}=\sqrt{13}$

Thus,

$e=\frac{\sqrt{13}}{2}$

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Example Question #192 : Conic Sections

Find the eccentricity of the following hyperbola:

$\frac{(y-12)^2}{100}-\frac{(x-1)^2}{300}=1$

Possible Answers:

$\frac{1}{2}$

$\frac{4}{3}$

Correct answer:

Explanation:

Recall the standard form of the equation of a vertical hyperbola:

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$ , where (h,k) is the center

For a horizontal hyperbola, use the following equation to find the eccentricity:

$e=\frac{c}{b}$ , where $c=\sqrt{a^b+b^2}$

For the given hyperbola,

$a=\sqrt{300}$ and b=10

$c=\sqrt{300+100}=\sqrt{400}=20$

Thus,

$e=\frac{20}{10}=2$

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Example Question #102 : Understand Features Of Hyperbolas And Ellipses

Find the eccentricity of the following hyperbola:

$\frac{(y-3)^2}{48}-\frac{(x-8)^2}{52}=1$

Possible Answers:

$\frac{\sqrt3}{6}$

$\frac{6\sqrt3}{5}$

$\frac{13}{12}$

$\frac{5\sqrt3}{6}$

Correct answer:

$\frac{5\sqrt3}{6}$

Explanation:

Recall the standard form of the equation of a vertical hyperbola:

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$ , where (h,k) is the center

For a horizontal hyperbola, use the following equation to find the eccentricity:

$e=\frac{c}{b}$ , where $c=\sqrt{a^b+b^2}$

For the given hyperbola,

$a=\sqrt{52}$ and $b=\sqrt{48}$

$c=\sqrt{48+52}=\sqrt{100}=10$

Thus,

$e=\frac{10}{\sqrt{48}}=\frac{10\sqrt{48}}{48}=\frac{5\sqrt{48}}{24}=\frac{20\sqrt3}{24}=\frac{5\sqrt3}{6}$

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Example Question #103 : Understand Features Of Hyperbolas And Ellipses

Find the eccentricity for the following hyperbola:

$\frac{(y+2)^2}{121}-\frac{(x-7)^2}{21}=1$

Possible Answers:

$\frac{121}{21}$

$\frac{12}{11}$

$\frac{\sqrt{11}}{142}$

$\frac{\sqrt{142}}{11}$

Correct answer:

$\frac{\sqrt{142}}{11}$

Explanation:

Recall the standard form of the equation of a vertical hyperbola:

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$ , where (h,k) is the center

For a horizontal hyperbola, use the following equation to find the eccentricity:

$e=\frac{c}{b}$ , where $c=\sqrt{a^b+b^2}$

For the given hyperbola,

$a=\sqrt{21}$ and b=11

$c=\sqrt{121+21}=\sqrt{142}$

Thus,

$e=\frac{\sqrt{142}}{11}$

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Example Question #101 : Hyperbolas And Ellipses

Find the eccentricity for the following hyperbola:

$\frac{(y-5)^2}{10}-\frac{(x+1)^2}{9}=1$

Possible Answers:

$\frac{10}{9}$

$\frac{3}{10}$

$\frac{\sqrt{190}}{10}$

$\frac{\sqrt10}{3}$

Correct answer:

$\frac{\sqrt{190}}{10}$

Explanation:

Recall the standard form of the equation of a vertical hyperbola:

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$ , where (h,k) is the center

For a horizontal hyperbola, use the following equation to find the eccentricity:

$e=\frac{c}{b}$ , where $c=\sqrt{a^b+b^2}$

For the given hyperbola,

a=3 and $b=\sqrt{10}$

$c=\sqrt{9+10}=\sqrt{19}$

Thus,

$e=\frac{\sqrt{19}}{\sqrt{10}}=\frac{\sqrt{190}}{10}$

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Example Question #1781 : Pre Calculus

Find the eccentricity of the following hyperbola:

91y^2-9x^2+182y+90x-953=0

Possible Answers:

$\frac{3\sqrt{10}}{8}$

$\frac{3}{10}$

$\frac{10}{3}$

$\frac{\sqrt{3}}{10}$

Correct answer:

$\frac{10}{3}$

Explanation:

Recall the standard form of the equation of a vertical hyperbola:

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$ , where (h,k) is the center

We need to put the given equation into the standard form of the equation.

Start by grouping like terms together.

91y^2+182y-9x^2+90x=953

Factor out from the terms and from the terms.

91(y^2+2y)-9(x^2-10x)=953

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

91(y^2+2y+1)-9(x^2-10x+25)=819

Divide both sides by 819 .

$\frac{y^2+2y+1}{9}-\frac{x^2-10x+25}{91}=1$

Factor both terms to get the standard equation.

$\frac{(y+1)^2}{9}-\frac{(x-5)^2}{91}=1$

For a horizontal hyperbola, use the following equation to find the eccentricity:

$e=\frac{c}{b}$ , where $c=\sqrt{a^b+b^2}$

For the given hyperbola,

$a=\sqrt{91}$ and b=3

$c=\sqrt{91+9}=\sqrt{100}=10$

Thus,

$e=\frac{10}{3}$

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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 #102 : Hyperbolas And Ellipses

Find the foci of the hyperbola with the following equation:

$\frac{(x-3)^2}{45}-\frac{(y-2)^2}{4}=1$

Possible Answers:

(10, 2)(-4, 2)

(3, 2)(10, 2)

(5, 2)(1, 2)

(3, 9)(3, -5)

Correct answer:

(10, 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) .

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 (3,2) .

Next, find .

$c=\sqrt{45+4}=\sqrt{49}=7$

The foci are then located at (10, 2) and (-4, 2) .

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Precalculus : Hyperbolas and Ellipses

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1775 : Pre Calculus

Example Question #101 : Understand Features Of Hyperbolas And Ellipses

Example Question #192 : Conic Sections

Example Question #102 : Understand Features Of Hyperbolas And Ellipses

Example Question #103 : Understand Features Of Hyperbolas And Ellipses

Example Question #101 : Hyperbolas And Ellipses

Example Question #1781 : Pre Calculus

Example Question #41 : Graphs

Example Question #201 : Conic Sections

Example Question #102 : Hyperbolas And Ellipses

All Precalculus Resources

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