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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

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Example Questions

Example Question #1771 : Pre Calculus

Find the eccentricity of the following hyperbola:

$\frac{(x+2)^2}{9}-\frac{(y-1)^2}{25}=1$

Possible Answers:

$\frac{5}{3}$

$\frac{3\sqrt{34}}{3}$

$\frac{9}{25}$

$\frac{\sqrt{34}}{3}$

Correct answer:

$\frac{\sqrt{34}}{3}$

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

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=3 and b=5

$c=\sqrt{9+25}=\sqrt{34}$

Thus,

$e=\frac{\sqrt{34}}{3}$

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Example Question #1772 : Pre Calculus

Find the eccentricity of the following hyperbola:

$\frac{(x+2)^2}{13}-\frac{(y-9)^2}{12}=1$

Possible Answers:

$\frac{5\sqrt{13}}{12}$

$\frac{5\sqrt{13}}{13}$

$\frac{13}{12}$

$\frac{5}{13}$

Correct answer:

$\frac{5\sqrt{13}}{13}$

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

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,

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

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

Thus,

$e=\frac{5}{\sqrt{13}}=\frac{5\sqrt{13}}{13}$

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Example Question #95 : Hyperbolas And Ellipses

Find the eccentricity of the following hyperbola:

$\frac{(x-3)^2}{21}-\frac{(y+9)^2}{28}=1$

Possible Answers:

$\frac{3}{4}$

$\frac{4}{3}$

$\frac{\sqrt3}{21}$

$\frac{\sqrt{21}}{3}$

Correct answer:

$\frac{\sqrt{21}}{3}$

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

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,

$a=\sqrt{21}$ and $b=\sqrt{28}$

$c=\sqrt{21+28}=\sqrt{49}=7$

Thus,

$e=\frac{7}{\sqrt{21}}=\frac{7\sqrt{21}}{21}=\frac{\sqrt{21}}{3}$

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Example Question #191 : Conic Sections

Find the eccentricity of the following hyperbola:

$\frac{(x-3)^1}{100}-\frac{(y+4)^2}{21}=1$

Possible Answers:

$\frac{11}{10}$

$\frac{10}{11}$

$\frac{100}{11}$

$\frac{121}{100}$

Correct answer:

$\frac{11}{10}$

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

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=10 and $b=\sqrt{21}$

$c=\sqrt{100+21}=\sqrt{121}=11$

Thus,

$e=\frac{11}{10}$

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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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Precalculus : Pre-Calculus

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1771 : Pre Calculus

Example Question #1772 : Pre Calculus

Example Question #95 : Hyperbolas And Ellipses

Example Question #191 : Conic Sections

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

All Precalculus Resources

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