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Precalculus : Hyperbolas and Ellipses

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Example Questions

Example Question #1 : Hyperbolas And Ellipses

Consider the following equation:

9x^2 - 18x + 100y^2 + 600y = -9

Classify the equation by its form when graphed.

Possible Answers:

A hyperbola with foci at (3, -1) and (1, 3)

An ellipse centered at (1, -3)

A circle centered at (3, -1)

A hyperbola with foci at (10, 3) and (-10, -3)

An ellipse centered at (10, 3)

Correct answer:

An ellipse centered at (1, -3)

Explanation:

Start with the equation

9x^2 - 18x + 100y^2 + 600y = -9

It would be helpful if we "complete the square" for both the x and y polynomials.

So by adding 9 and 900 we transform the equation into

And then by dividing 900, we see that the outcome is:

$\frac{(x-1)^2}{100} + \frac{(y+3)^2}{9} = 1$

The standard equation for an ellipse is

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

With a center at (h, k)

Therefore this is the equation of an ellipse, with center (1, -3)

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Example Question #2 : Hyperbolas And Ellipses

Express the following equation for an ellipse in standard form:

4(x-4)^2+9(x-7)^2=36

Possible Answers:

$\frac{(x-4)^2}{9}+\frac{(y-7)^2}{4}=1$

$\frac{(x-4)^2}{9}+\frac{(y-7)^2}{4}=36$

4(x-4)^2+9(y-7)^2=1

$\frac{4(x-4)^2}{36}+\frac{9(y-7)^2}{36}=36$

(x-4)^2+(y-7)^2=1

Correct answer:

$\frac{(x-4)^2}{9}+\frac{(y-7)^2}{4}=1$

Explanation:

Remember that the equation for an ellipse in standard form looks like the following:

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

Where the point (h,k) gives the center of the ellipse, a is half the length of its axis in the x direction, and b is half the length of its axis in the y direction. We can see that this form has a 1 on the right side of the equation, so let's start by dividing both sides of our equation by 36 to get a 1 on the right side:

4(x-4)^2+9(x-7)^2=36

$\frac{4(x-4)^2+9(x-7)^2}{36}=\frac{36}{36}$

$\frac{4(x-4)^2}{36}+\frac{9(x-7)^2}{36}=1$

Now we can simplify the fractions on the right side of the equation, which gives us the equation for our ellipse in standard form:

$\frac{(x-4)^2}{9}+\frac{(y-7)^2}{4}=1$

This ellipse would have its center at (4,7), would be 6 units wide in the x direction, and 4 units wide in the y direction, because a^2=9 so a=3 , and b^2=4 so b=2 .

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Example Question #3 : Hyperbolas And Ellipses

Which of the following is an equation for an ellipse written in standard form?

Possible Answers:

(y+2)^2+(x-4)^2=81

$\frac{(y-9)^2}{25}-\frac{(x-10)^2}{9}=1$

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

$\frac{(x+3)^2}{16}-\frac{(y+4)^2}{25}=1$

$\frac{(y-3)^2}{4}+\frac{(x+5)^2}{16}=1$

Correct answer:

$\frac{(y-3)^2}{4}+\frac{(x+5)^2}{16}=1$

Explanation:

Remember that in order for the equation of an ellipse to be written in standard form, it must be written in one of the following two ways:

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

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

Where the point (h,k) gives the center of the ellipse, a is half the length of the axis for which it is the denominator, and b is half the length of the axis for which it is the denominator. Looking at our answer choices, we need one that involves addition, has different denominators under the x and y terms, and is equal to 1 on the right side of the equation. We can see that out of all the answer choices, the following is the only one that satisfies those requirements of standard form:

$\frac{(y-3)^2}{4}+\frac{(x+5)^2}{16}=1$

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Example Question #1 : Hyperbolas And Ellipses

What is the equation of an ellipse in standard form with the point (3,6) as its center, an -axis length of 18cm , and a -axis length of 10cm .

Possible Answers:

$\frac{(x-3)^{2}}{81}+\frac{(y-6)^{2}}{25}=1$

$\frac{(x-3)^{2}}{18}+\frac{(y-6)^{2}}{10}=1$

$\frac{(x-3)^{2}}{10}+\frac{(y-6)^{2}}{18}=1$

$\frac{(x-3)^{2}}{25}+\frac{(y-6)^{2}}{81}=1$

$\frac{(x-6)^{2}}{81}+\frac{(y-3)^{2}}{25}=1$

Correct answer:

$\frac{(x-3)^{2}}{81}+\frac{(y-6)^{2}}{25}=1$

Explanation:

The equation of an ellipse in standard form is

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

where (h,k) is the center point, is half the length of its axis in the direction, and is half the length of the axis in the direction.

In this instance, the center point (h,k)= (3,6) , $a=\frac{1}2{(18)}=9$ , and $b=\frac{1}{2}(10)=5$ . Therefore:

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

$\frac{(x-3)^{2}}{9^{2}}+\frac{(y-6)^{2}}{5^{2}}=1$

$\frac{(x-3)^{2}}{81}+\frac{(y-6)^{2}}{25}=1$

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Example Question #3 : Hyperbolas And Ellipses

What is the equation of a circle with radius of and center of (0, 0) ?

Possible Answers:

(x - 1)^2 + (y + 1)^2 = 12

(x - 2)^2 + (y + 4)^2 = 144

5x^2 + 4x + 2 = 10

x^2 + y^2 = 144

x^2 + y = 12

Correct answer:

x^2 + y^2 = 144

Explanation:

Recall that the general equation of a circle is:

(x - h)^2 + (y - k)^2 = r^2 .

The key feature here being that both the and terms are squared.

Thus, we can narrow down the choices without this feature.

Next, consider the center. Since the center here is (0,0) , we should not have any middle terms.

The only choice that fits that description is x^2 + y^2 = 144 .

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Example Question #2 : Hyperbolas And Ellipses

Write the equation for an ellipse with center (1, 4) , foci $(1 \pm \sqrt{14}, 4)$ and a major axis with length 14.

Possible Answers:

$\frac{(x-4)^2 }{25 } + \frac{(y-1)^2 }{ 49} = 1$

$\frac{(x-1)^2 }{ 49 } + \frac{(y-4)^2 }{35 } = 1$

$\frac{(x-1)^2 }{ 35 } + \frac{(y-4)^2 }{49 } = 1$

$\frac{(x+1)^2}{49 } + \frac{(y+4)^2}{35 } =1$

$\frac{(x-1)^2 }{ 196} + \frac{(y-4)^2 }{ 182} = 1$

Correct answer:

$\frac{(x-1)^2 }{ 49 } + \frac{(y-4)^2 }{35 } = 1$

Explanation:

The general equation for an ellipse is $\frac{(x-h)^2 }{a^2 } + \frac{(y-k)^2 }{b^2 } = 1$ , although if we consider a to be half the length of the major axis, a and b might switch depending on if the longer major axis is horizontal or vertical. This general equation has (h,k) as the center, a as the length of half the major axis, and b as the length of half the minor axis.

Because the center of this ellipse is at (1, 4) and the foci are at $(1 \pm \sqrt{14} , 4 )$ , we can see that the foci are $\sqrt{14}$ away from the center, and they are on the horizontal axis. This means that the horizontal axis is the major axis, the one with length 14. Having a length of 14 means that half is 7, so a = 7 . Since the foci are $\sqrt{14}$ away from the center, we know that $c = \sqrt{14}$ . We can solve for b using the equation a^2 - c^2 = b^2 :

$7^2 - \sqrt{14}^2 = b^2$

49 - 14 = b^2

35 = b^2 that's really as far as we need to solve.

Putting all this information into the equation gives:

$\frac{(x-1)^2 }{ 49} + \frac{(y - 4)^2 }{35 } = 1$

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Example Question #1 : Understand Features Of Hyperbolas And Ellipses

Find the equation of an ellipse centered at the origin if the major axis is parallel to the x-axis and has a length of units and the minor axis has a length of units.

Possible Answers:

$\small \frac{x^2}{16}+\frac{y^2}{4}=1$

$\small \frac{x^2}{64}+\frac{y^2}{16}=1$

$\small \small \frac{x^2}{16}+\frac{y^2}{64}=1$

$\small \small \frac{x^2}{4}+\frac{y^2}{16}=1$

Correct answer:

$\small \frac{x^2}{16}+\frac{y^2}{4}=1$

Explanation:

The formula for an ellipse centered at the point (h, k) with a horizontal major axis (ie: parallel to the x-axis) has the formula

$\small \frac{(x-h)^2}{[\frac{a}{2}]^2}+\frac{(y-k)^2}{[\frac{b}{2}]^2}=1$

where $\small a$ and $\small b$ are the lengths of the major and minor axes respectively.

Since the origin is at $\small (0,0)$ and $\small a=8$ and $\small b=4$ in this problem, the equation is

$\small \frac{(x-0)^2}{[\frac{8}{2}]^2}+\frac{(y-0)^2}{[\frac{4}{2}]^2}=1$

$\small \frac{x^2}{16}+\frac{y^2}{4}=1$

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Example Question #4 : Hyperbolas And Ellipses

The equation of an ellipse, , is $\frac{(x+3)^2}{9}+\frac{(y-7)^2}{36}=1$ . Which of the following is the correct center and foci of this ellipse?

Possible Answers:

Center= (3,6) , Foci= (3,1) and (3,11)

Center= (-3,7) , Foci= (-8,7) and (2,7)

Center= (-3,7) , Foci= (-3,2) and (-3,12)

Center= (3,-7) , Foci= (3,-2) and

Center= (3,-7) , Foci= (-2,7) and (8,7)

Correct answer:

Center= (-3,7) , Foci= (-3,2) and (-3,12)

Explanation:

Because our equation is already in the format

$\frac{(x-h)^2}{w^2}+\frac{(y-k)^2}{v^2}=1$ ,

we do not have to manipulate the equation. The center of any ellipse in this form will always be . So in this case, our center will be (-3,7) . To find the foci of the ellipse, we must use the equation b^2+c^2=a^2 , where a^2 is the greater of the two denominators in our equation ( w^2 and v^2 ), b^2 is the lesser and is the distance from the center to the foci.

We know that a^2=36 and b^2=9 .

By using a^2-b^2=c^2 , we see that c^2=25 , so c=5 .

We now know that the two foci are going to be units in either direction of the center along the greater axis.

Because the greater denominator is under our term containing , our ellipse will have its greater axis going vertically, rather than horizontally. Therefore, our foci will be units above and below our center, at (-3,2) and (-3,12) .

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Example Question #1 : Understand Features Of Hyperbolas And Ellipses

Find the center of the ellipse with the following equation:

$\frac{(x-1)^2}{36}+\frac{(y+2)^2}{9}=1$

Possible Answers:

(1, -2)

(-1, 2)

(6, 3)

(-6, -3)

Correct answer:

(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.

For the equation given in the question, h=1 and k=-2 .

The center of the ellipse is at (1, -2)

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Example Question #1 : Hyperbolas And Ellipses

Find the center of the ellipse with the following equation:

$\frac{(x+9)^2}{2}+\frac{(y-4)^2}{3}=1$

Possible Answers:

$(\sqrt2, \sqrt3)$

(-9, 4)

(9, -4)

$(-\sqrt2, -\sqrt3)$

Correct answer:

(-9, 4)

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=-9 and k=4 .

The center of the ellipse is at (-9, 4) .

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Precalculus : Hyperbolas and Ellipses

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