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Precalculus : Conic Sections

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Example Questions

Example Question #1763 : Pre Calculus

Find the center of the hyperbola with the following equation:

25x^2-4y^2+32y-164=0

Possible Answers:

(5, 0)

(0, 4)

(2, 5)

(4, 0)

Correct answer:

(0, 4)

Explanation:

Recall the standard form of the equation of a hyperbola can come in two forms:

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

In both cases, the center of the hyperbola is located at (h, k) .

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

Factor out from the terms.

25x^2-4y^2+32y-164=0

25x^2-4(y^2-8y)-164=0

Now complete the square. Remember to add the same amount to both sides of the equation!

25x^2-4(y^2-8y+16)-164=-64

Add 164 to both sides of the equation.

25x^2-4(y^2-8y+16)=100

Factor the square portion of the equation.

25x^2-4(y-4)^2=100

Divide both sides by 100 to get the standard form of the equation of a hyperbola..

$\frac{x^2}{4}-\frac{(y-4)^2}{25}=1$

For the hyperbola in question, h=0 and k=4 , so the center is at (0, 4) .

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Example Question #88 : Understand Features Of Hyperbolas And Ellipses

Find the center of the hyperbola with the following equation:

4y^2-3x^2-48y-6x+129=0

Possible Answers:

(3, 4)

(6, -1)

(-1, 6)

(4, -3)

Correct answer:

(-1, 6)

Explanation:

Recall the standard form of the equation of a hyperbola can come in two forms:

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

In both cases, the center of the hyperbola is located at (h, k) .

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

Group the terms together and the terms together.

4y^2-3x^2-48y-6x+129=0

4y^2-48y-3x^2-6x+129=0

Factor out from the terms and from the terms.

4(y^2-12y)-3(x^2+2x)+129=0

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

4(y^2-12y+36)-3(x^2+2x+1)+129=141

Subtract 129 from both sides of the equation.

4(y^2-12y+36)-3(x^2+2x+1)=12

Factor the square portions of the equation.

4(y-6)^2-3(x+1)^2=12

Divide both sides by to get the standard form of the equation of a hyperbola.

$\frac{(y-6)^2}{3}-\frac{(x+1)^2}{4}=1$

For the hyperbola in question, h=-1 and k=6 , so the center is at (-1, 6) .

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Example Question #181 : Conic Sections

Find the vertices of the hyperbola with the following equation:

4y^2-25x^2-56y-150x-129=0

Possible Answers:

(-1, 7)(-5, 7)

(9, 1)(-1, 9)

(-3, 12)(-3, 2)

(12, 3)(12, -3)

Correct answer:

(-3, 12)(-3, 2)

Explanation:

Recall the standard form of the equation of a hyperbola can come in two forms:

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

In both cases, the center of the hyperbola is located at (h, k) .

For the hyperbola with the equation $\frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1$ , the vertices are located at (h-a, k)(h+a, k) .

For the hyperbola with the equation $\frac{(y-k^2)}{b^2}-\frac{(x-h)^2}{a^2}=1$ , the vertices are located at (h, k+b)(h, k-b) .

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

Group the terms together and the terms together.

4y^2-25x^2-56y-150x-129=0

4y^2-56y-25x^2-150x-129=0

Factor out from the terms and from the terms.

4(y^2-14y)-25(x^2+6x)-129=0

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

4(y^2-14y+49)-25(x^2+6x+9)-129=-29

Add 129 to both sides of the equation.

4(y^2-14y+49)-25(x^2+6x+9)=100

Factor the square portions of the equation.

4(y-7)^2-25(x+3)^2=100

Divide both sides by 100 to get the standard form for the equation of a hyperbola.

$\frac{(y-7)^2}{25}-\frac{(x+3)^2}{4}=1$

For the hyperbola in question, the center is located at (-3, 7) and b=5 . The vertices must be at (-3, 12) and (-3, 2) .

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Example Question #92 : Understand Features Of Hyperbolas And Ellipses

Find the vertices of the hyperbola with the following equation:

100y^2-225x^2-2000y+450x-12725=0

Possible Answers:

(11, 10)(-9, 10)

(1, 25)(1, -5)

(16, 10)(-14, 10)

(1, 20)(1, 0)

Correct answer:

(1, 25)(1, -5)

Explanation:

Recall the standard form of the equation of a hyperbola can come in two forms:

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

In both cases, the center of the hyperbola is located at (h, k) .

For the hyperbola with the equation $\frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1$ , the vertices are located at (h-a, k)(h+a, k) .

For the hyperbola with the equation $\frac{(y-k^2)}{b^2}-\frac{(x-h)^2}{a^2}=1$ , the vertices are located at (h, k+b)(h, k-b) .

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

Group the terms together and the terms together.

100y^2-225x^2-2000y+450x-12725=0

100y^2-2000y-225x^2+450x-12725=0

Factor out 100 from the terms and from the terms.

100(y^2-20y)-225(x^2-2x)-12725=0

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

100(y^2-20y+100)-225(x^2-2x+1)-12725=9775

Add 12725 to both sides of the equation.

100(y^2-20y+100)-225(x^2-2x+1)=22500

Factor the square portions of the equation.

100(y-10)^2-225(x-1)^2=22500

Divide both sides by 22500 to get the standard form for the equation of a hyperbola.

$\frac{(y-10)^2}{225}-\frac{(x-1)^2}{100}=1$

For the hyperbola in question, the center is located at (1, 10) and b=15 . The vertices must be at (1, 25) and (1, -5) .

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Example Question #93 : Understand Features Of Hyperbolas And Ellipses

Find the endpoints of the conjugate axis of the hyperbola with the following equation:

4x^2-16y^2-16x+64y-112=0

Possible Answers:

(2, 4)(2, 0)

(0, 2)(4, 2)

(4, 4)(0, 0)

(2, 4)(2, -4)

Correct answer:

(2, 4)(2, 0)

Explanation:

Recall the standard form of the equation of a hyperbola can come in two forms:

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

In both cases, the center of the hyperbola is located at (h, k) .

For the hyperbola with the equation $\frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1$ , the endpoints of the conjugate axis are located at (h, k+b)(h, k-b) .

For the hyperbola with the equation $\frac{(y-k^2)}{b^2}-\frac{(x-h)^2}{a^2}=1$ , the endpoints of the conjugate axis are located at (h-a, k)(h+a, k) .

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

Group the terms together and the terms together.

4x^2-16y^2-16x+64y-112=0

4x^2-16x-16y^2+64y-112=0

Factor out from the terms and from the terms.

4(x^2-4x)-16(y^2-4y)-112=0

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

4(x^2-4x+4)-16(y^2-4y+4)-112=-48

Add 112 to both sides of the equation.

4(x^2-4x+4)-16(y^2-4y+4)=64

Factor the squares.

4(x-2)^2-16(y-2)^2=64

Divide both sides by to get the standard form for the equation of a hyperbola.

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

For the hyperbola in question, the center is located at (2,2) and b=2 . The endpoints of the conjugate axis must be at (2, 0) and (2, 4) .

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Example Question #94 : Understand Features Of Hyperbolas And Ellipses

Find the endpoints of the conjugate axis of the hyperbola with the following equation:

9y^2-4x^2-216y+88x+776=0

Possible Answers:

(12, 14)(12, 8)

(11, 13)(11, 9)

(12, 9)(12, 13)

(14,12)(8, 12)

Correct answer:

(14,12)(8, 12)

Explanation:

Recall the standard form of the equation of a hyperbola can come in two forms:

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

In both cases, the center of the hyperbola is located at (h, k) .

For the hyperbola with the equation $\frac{(x-h)^2}{a^2}-\frac{(y-k^2)}{b^2}=1$ , the endpoints of the conjugate axis are located at (h, k+b)(h, k-b) .

For the hyperbola with the equation $\frac{(y-k^2)}{b^2}-\frac{(x-h)^2}{a^2}=1$ , the endpoints of the conjugate axis are located at (h-a, k)(h+a, k) .

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

Group the terms together and the terms together.

9y^2-4x^2-216y+88x+776=0

9y^2-216y-4x^2+88x+776=0

Factor out from the terms and from the terms.

9(y^2-24y)-4(x^2-22x)+776=0

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

9(y^2-24y+144)-4(x^2-22x+121)+776=812

Subtract 776 from both sides of the equation.

9(y^2-24y+144)-4(x^2-22x+121)=36

Factor the squares.

9(y-12)^2-4(x-11)^2=36

Divide both sides by to get the standard form for the equation of a hyperbola.

$\frac{(y-12)^2}{4}-\frac{(x-11)^2}{9}=1$

For the hyperbola in question, the center is located at (11, 12) and a=3 . The endpoints of the conjugate axis must be at (14, 12) and (8, 12) .

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Example Question #95 : Understand Features Of Hyperbolas And Ellipses

Find the eccentricity of the following hyperbola: $\frac{x^2}{4}-\frac{y^2}{25}=1$

Possible Answers:

$\sqrt{34}$

$\frac{\sqrt{29}}{5}$

$\frac{\sqrt{41}}{5}$

$\frac{\sqrt{41}}{2}$

$\frac{\sqrt{29}}{2}$

Correct answer:

$\frac{\sqrt{29}}{2}$

Explanation:

In order to find the eccentricity of $\frac{x^2}{4}-\frac{y^2}{25}=1$ , first determine the values of and from the standard form of the hyperbola:

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

a^2=4

a=2

b^2=25

b=5

Use the following formula to calculate eccentricity. The eccentricity of a hyperbola should always be greater than 1.

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

Substitute and solve for eccentricity.

$e=\frac{\sqrt{2^2+5^2}}{2}=\frac{\sqrt{29}}{2}$

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Example Question #96 : Understand Features Of Hyperbolas And Ellipses

Find the eccentricity of the hyperbola with the following equation:

$\frac{x^2}{144}-\frac{y^2}{25}=1$

Possible Answers:

$\frac{25}{144}$

$\frac{144}{25}$

$\frac{13}{12}$

$\frac{12}{13}$

Correct answer:

$\frac{13}{12}$

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.

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=12 and b=5

$c=\sqrt{144+25}=\sqrt{169}=13$

Thus,

$e=\frac{13}{12}$

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Example Question #91 : Understand Features Of Hyperbolas And Ellipses

Find the eccentricity of the following hyperbola:

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

Possible Answers:

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

$\frac{5}{3}$

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

$\frac{9}{25}$

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 #92 : Understand Features Of Hyperbolas And Ellipses

Find the eccentricity of the following hyperbola:

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

Possible Answers:

$\frac{5}{13}$

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

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

$\frac{13}{12}$

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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Precalculus : Conic Sections

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1763 : Pre Calculus

Example Question #88 : Understand Features Of Hyperbolas And Ellipses

Example Question #181 : Conic Sections

Example Question #92 : Understand Features Of Hyperbolas And Ellipses

Example Question #93 : Understand Features Of Hyperbolas And Ellipses

Example Question #94 : Understand Features Of Hyperbolas And Ellipses

Example Question #95 : Understand Features Of Hyperbolas And Ellipses

Example Question #96 : Understand Features Of Hyperbolas And Ellipses

Example Question #91 : Understand Features Of Hyperbolas And Ellipses

Example Question #92 : Understand Features Of Hyperbolas And Ellipses

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