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Precalculus : Hyperbolas and Ellipses

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Example Questions

Example Question #11 : Understand Features Of Hyperbolas And Ellipses

Find the center of the ellipse with the following equation:

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

Possible Answers:

(-4, 2)

$(\sqrt2, 2\sqrt2)$

(4, -2)

$(-\sqrt2, -2\sqrt2)$

Correct answer:

(-4, 2)

Explanation:

Recall that the standard form of the equation of an ellipse is

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

For the equation given in the question, h=-4 and k=2 .

The center of the ellipse is at (-4, 2) .

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Example Question #11 : Hyperbolas And Ellipses

Find the center of the ellipse with the following equation:

25x^2+9y^2+200x-36y+211=0

Possible Answers:

(3, 5)

(-4, 2)

(25, 9)

(2, 4)

Correct answer:

(-4, 2)

Explanation:

Start by putting the equation back into the standard equation of the ellipse:

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

Group the terms and terms together.

25x^2+9y^2+200x-36y+211=0

25x^2+200x+9y^2-36y+211=0

Factor out a from the terms, and a from the terms.

25(x^2+8x)+9(y^2-4y)+211=0

Now, complete the square. Remember to add the same amounts on both sides of the equation.

25(x^2+8x+16)+9(y^2-4y+4)+211=436

25(x^2+8x+16)+9(y^2-4y+4)=225

Now, divide both sides by 225 .

$\frac{x^2+8x+16}{9}+\frac{y^2-4y+4}{25}=1$

Finally, factor the equations to get the standard form of the equation for an ellipse.

$\frac{(x+4)^2}{9}+\frac{(y-2)^2}{25}=1$

Since h=-4 and k=2 , the center for this ellipse is (-4, 2) .

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Example Question #12 : Hyperbolas And Ellipses

Find the center of the ellipse with the following equation:

4x^2+6y^2-40x-36y+130=0

Possible Answers:

(4, 6)

(-3, 5)

(5, 3)

(40, 36)

Correct answer:

(5, 3)

Explanation:

Start by putting the equation back into the standard equation of the ellipse:

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

Group the terms and terms together.

4x^2+6y^2-40x-36y+130=0

4x^2-40x+6y^2-36y+130=0

Factor out a from the terms and a from the terms.

4(x^2-10x)+6(y^2-6y)+130=0

Now, complete the squares. Make sure you add the same amount on both sides!

4(x^2-10x+25)+6(y^2-6y+9)+130=154

Subtract 130 from both sides.

4(x^2-10x+25)+6(y^2-6y+9)=24

Now, divide both sides by .

$\frac{x^2-10x+25}{6}+\frac{y^2-6y+9}{4}=1$

Finally, factor the terms to get the standard form of the equation of an ellipse.

$\frac{(x-5)^2}{6}+\frac{(y-3)^2}{4}=1$

Since h=5 and k=3 , the center of the ellipse is (5, 3) .

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Example Question #11 : Understand Features Of Hyperbolas And Ellipses

Find the center of the ellipse with the following equation:

9x^2+8y^2-18x-112y+329=0

Possible Answers:

(6, 12)

(9, 8)

(1, 7)

(-1, -7)

Correct answer:

(1, 7)

Explanation:

Start by putting the equation back into the standard equation of the ellipse:

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

Group the terms and terms together.

9x^2+8y^2-18x-112y+329=0

9x^2-18x+8y^2-112y+329=0

Factor out a from the terms and a from the terms.

9(x^2-2x)+8(y^2-14y)+329=0

Now, complete the squares. Remember to add the same amount on both sides!

9(x^2-2x+1)+8(y^2-14y+49)+329=401

Subtract 329 from both sides.

9(x^2-2x+1)+8(y^2-14y+49)=72

Divide both sides by .

$\frac{x^2-2x+1}{8}+\frac{y^2-14y+49}{9}=1$

Finally, factor the terms to get the standard form of the equation of an ellipse.

$\frac{(x-1)^2}{8}+\frac{(y-7)^2}{9}=1$

Since h=1 and k=7 , (1, 7) is the center of this ellipse.

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Example Question #101 : Conic Sections

Find the foci of an ellipse with the following equation:

$\frac{(x-5)^2}{25}+\frac{(y-1)^2}{169}=1$

Possible Answers:

(5, 13), (5, -11)

(17, 1), (-7, 1)

(169, 25), (25, 169)

(-5, -13), (-5, 11)

Correct answer:

(5, 13), (5, -11)

Explanation:

Recall that the standard form of the equation of an ellipse is

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

When a^2>b^2 , the major axis will lie on the -axis and be horizontal. When b^2>a^2 , the major axis will lie on the -axis and be vertical.

Recall also that the distance from the center to a focus, , is given by the equation $c=\sqrt{a^2-b^2}$ when a>b , and the equation is $c=\sqrt{b^2-a^2}$ when b>a .

When the major axis follows the -axis, the points for the foci are (h+c, k) and (h-c, k) .

When the major axis follows the -axis, the points for the foci are (h, k+c) and (h, k-c) .

For the given equation, the center is at (5, 1) . Since b^2>a^2 , the major-axis is vertical.

Plug in the values to solve for .

$c=\sqrt{169-25}=\sqrt{144}=12$

Now, add to the y-coordinate of the center to get one focus. Subtract from the y-coordinate of the center to get the other focus point.

The foci for the ellipse is then (5, 13) and (5, -11) .

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Example Question #106 : Conic Sections

Find the foci of the ellipse with the following equation:

5x^2+6y^2+50x+72y+311=0

Possible Answers:

(-5, -4), (-5, -7)

(-5, 2), (-5, 1)

(6, 6)(7, 6)

(-6, -6), (-4, -6)

Correct answer:

(-6, -6), (-4, -6)

Explanation:

Recall that the standard form of the equation of an ellipse is

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

When a^2>b^2 , the major axis will lie on the -axis and be horizontal. When b^2>a^2 , the major axis will lie on the -axis and be vertical.

Recall also that the distance from the center to a focus, , is given by the equation $c=\sqrt{a^2-b^2}$ when a>b , and the equation is $c=\sqrt{b^2-a^2}$ when b>a .

When the major axis follows the -axis, the points for the foci are (h+c, k) and (h-c, k) .

When the major axis follows the -axis, the points for the foci are (h, k+c) and (h, k-c) .

Start by putting the equation into the standard form of the equation of an ellipse.

Group the and terms together.

5x^2+6y^2+50x+72y+311=0

5x^2+50x+6y^2+72y+311=0

Now, factor out a from the terms and a from the terms.

5(x^2+10x)+6(y^2+12y)+311=0

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

5(x^2+10x+25)+6(y^2+12y+36)+311=341

Subtract 311 from both sides of the equation.

5(x^2+10x+25)+6(y^2+12y+36)=30

Divide both sides by .

$\frac{x^2+10x+25}{6}+\frac{y^2+12y+36}{5}=1$

Factor both terms to get the standard form of the equation of an ellipse.

$\frac{(x+5)^2}{6}+\frac{(y+6)^2}{5}=1$

Now, the center for this ellipse is (-5, -6) and its major axis is horizontal.

Next, solve for .

$c=\sqrt{6-5}=\sqrt1=1$

The foci for this ellipse are then at (-6, -6) and (-4, -6) .

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Example Question #12 : Hyperbolas And Ellipses

Find the foci for the ellipse with the following equation:

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

Possible Answers:

$(\sqrt{119}, 0), (-\sqrt{119}, 0)$

(116, 0), (-116, 0)

(144, 25), (25, 144)

$(0, 2\sqrt{29}), (0, -2\sqrt{29})$

Correct answer:

$(\sqrt{119}, 0), (-\sqrt{119}, 0)$

Explanation:

Recall that the standard form of the equation of an ellipse is

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

When a^2>b^2 , the major axis will lie on the -axis and be horizontal. When b^2>a^2 , the major axis will lie on the -axis and be vertical.

Recall also that the distance from the center to a focus, , is given by the equation $c=\sqrt{a^2-b^2}$ when a>b , and the equation is $c=\sqrt{b^2-a^2}$ when b>a .

When the major axis follows the -axis, the points for the foci are (h+c, k) and (h-c, k) .

When the major axis follows the -axis, the points for the foci are (h, k+c) and (h, k-c) .

For the given equation, the center is at (0, 0) . Since a^2>b^2 , the major-axis is horizontal.

Plug in the values to solve for .

$c=\sqrt{144-25}=\sqrt{119}$

The foci are then at the points $(\sqrt{119}, 0)$ and $(-\sqrt{119}, 0)$ .

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Example Question #1691 : Pre Calculus

Find the foci of the ellipse with the following equation:

5x^2+9y^2-30x-36y+36=0

Possible Answers:

(3, 0), (3, 4)

(-5, -2)(-1, -2)

(5, 1), (8, 1)

(5, 2), (1, 2)

Correct answer:

(5, 2), (1, 2)

Explanation:

Recall that the standard form of the equation of an ellipse is

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

When a^2>b^2 , the major axis will lie on the -axis and be horizontal. When b^2>a^2 , the major axis will lie on the -axis and be vertical.

Recall also that the distance from the center to a focus, , is given by the equation $c=\sqrt{a^2-b^2}$ when a>b , and the equation is $c=\sqrt{b^2-a^2}$ when b>a .

When the major axis follows the -axis, the points for the foci are (h+c, k) and (h-c, k) .

When the major axis follows the -axis, the points for the foci are (h, k+c) and (h, k-c) .

Start by putting the equation into the standard form of the equation of an ellipse.

Group the and terms together.

5x^2+9y^2-30x-36y+36=0

5x^2-30x+9y^2-36y+36=0

Factor out a from the terms and a from the terms.

5(x^2-6x)+9(y^2-4y)+36=0

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

5(x^2-6x+9)+9(y^2-4y+4)+36=81

Subtract from both sides.

5(x^2-6x+9)+9(y^2-4y+4)=45

Divide both sides by .

$\frac{x^2-6x+9}{9}+\frac{y^2-4y+4}{5}=1$

Now, factor both terms to get the standard form of the equation of an ellipse.

$\frac{(x-3)^2}{9}+\frac{(y-2)^2}{5}=1$

The center of the ellipse is (3, 2) . Since a^2>b^2 , the major axis of this ellipse is horizontal.

Now, find the value of .

$c=\sqrt{9-5}=\sqrt4=2$

The foci of this ellipse are then (5, 2) and (1, 2) .

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Example Question #19 : Understand Features Of Hyperbolas And Ellipses

Find the foci of an ellipse with the following equation:

30x^2+5y^2+240x+80y+650=0

Possible Answers:

(-9, -8), (1, -8)

(-4, -13), (-4, -3)

(4, 3), (4, 13)

(-4, -8), (-4, -13)

Correct answer:

(-4, -13), (-4, -3)

Explanation:

Recall that the standard form of the equation of an ellipse is

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

When a^2>b^2 , the major axis will lie on the -axis and be horizontal. When b^2>a^2 , the major axis will lie on the -axis and be vertical.

Recall also that the distance from the center to a focus, , is given by the equation $c=\sqrt{a^2-b^2}$ when a>b , and the equation is $c=\sqrt{b^2-a^2}$ when b>a .

When the major axis follows the -axis, the points for the foci are (h+c, k) and (h-c, k) .

When the major axis follows the -axis, the points for the foci are (h, k+c) and (h, k-c) .

Start by putting the equation into the standard form of the equation of an ellipse.

Group the and terms together.

30x^2+5y^2+240x+80y+650=0

30x^2+240x+5y^2+80y+650=0

Factor out from the terms and a from the terms.

30(x^2+8x)+5(y^2+16y)+650=0

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

30(x^2+8x+16)+5(y^2+16y+64)+650=800

Subtract both sides by 650 .

30(x^2+8x+16)+5(y^2+16y+64)=150

Divide both sides by 150 .

$\frac{x^2+8x+16}{5}+\frac{y^2+16y+64}{30}=1$

Factor both terms to get the standard form of the equation of an ellipse.

$\frac{(x+4)^2}{5}+\frac{(y+8)^2}{30}=1$

The center of this ellipse is at (-4, -8) . Since b^2>a^2 , the major axis of this ellipse is vertical.

Now, solve for .

$c=\sqrt{30-5}=\sqrt{25}=5$

The foci for this ellipse are then (-4, -13) and (-4, -3) .

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Example Question #11 : Hyperbolas And Ellipses

Find the center and foci of the ellipse

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

Possible Answers:

Center: (0,2) ; Foci: $(\pm \sqrt{20}, 2 )$

Center: (0,2) ; Foci: $(0, 2 \pm \sqrt{20})$

Center: (0,-2) ; Foci: $(0, -2 \pm \sqrt{52})$

Center: (0,2) ; Foci: $(0, 2 \pm \sqrt{52})$

Center: (0, -2) ; Foci: $(\pm \sqrt{52}, -2)$

Correct answer:

Center: (0,2) ; Foci: $(0, 2 \pm \sqrt{20})$

Explanation:

The center of this ellipse is (0,2) . The number under (y-2)^2 is bigger than the number under x^2 , so the major axis goes up and down. The foci will also be on the major axis, so their x-coordinates will be 0, like the center.

To figure out the distance from the center to the foci, we can use the formula a^2 - c^2 = b^2 where a is half the major axis, b is half the minor axis, and c is the distance from the center to the foci.

In this case, $a = \sqrt{36} = 6$ and $b = \sqrt{16 } = 4$ :

36 - c^2 = 16 subtract 36 from both sides

-c^2 = -20 multiply both sides by -1

c^2 = 20 take the square root

$c = \sqrt{20}$

This means that since the center is (0,2) , the foci are located at $(0, 2+ \sqrt{20})$ and $(0, 2- \sqrt{20})$ .

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Precalculus : Hyperbolas and Ellipses

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #11 : Understand Features Of Hyperbolas And Ellipses

Example Question #11 : Hyperbolas And Ellipses

Example Question #12 : Hyperbolas And Ellipses

Example Question #11 : Understand Features Of Hyperbolas And Ellipses

Example Question #101 : Conic Sections

Example Question #106 : Conic Sections

Example Question #12 : Hyperbolas And Ellipses

Example Question #1691 : Pre Calculus

Example Question #19 : Understand Features Of Hyperbolas And Ellipses

Example Question #11 : Hyperbolas And Ellipses

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