Question

Find an equation for the conic that satisfies the given conditions. ellipse, foci (0, 3), (0, 7), vertices (0, 0), (0, 10)

Answer 1

An equation for the ellipse that satisfies the given conditions is  $\frac{x^2}{25}+\frac{(y-5)^2}{21}=1$

For given question,

We need to find an equation of ellipse having foci (0, 3), (0, 7), vertices (0, 0), (0, 10)

The x -coordinates of the vertices and foci are the same, so the major axis is parallel to the y -axis.

Thus, the equation of the ellipse will have the form

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

First, we identify the center,  (h, k) . 

The center is halfway between the vertices, (0, 0), (0, 10)

 Applying the midpoint formula, we have:

$\Rightarrow (h,k)=(\frac{0+0}{2} ,\frac{0+10}{2} )\\\\\Rightarrow (h,k)=(0,5)$

Next, we find  a² . 

The length of the major axis,  2a , is bounded by the vertices.

We solve for  'a'  by finding the distance between the y-coordinates of the vertices.

⇒ 2a = 10 - 0

⇒ 2a = 10

⇒ a = 5

⇒ a² = 25

Now we find  c² . 

The foci are given by  (h, k ± c)

So,  (h, k - c) = (0, 3) and  (h, k + c) = (0, 7)

k - c = 3

k + c = 7

After solving above system of equations we have,

c = 2 and k = 5

So, c² = 4

Next, we solve for  b²  using the equation  c² = a² - b²

⇒ 4 = 25 - b²

⇒ b² = 25 - 4

⇒ b² = 21

Now, we substitute the values found for  h, k, a², and  b²  into the standard form equation for an ellipse:

$\Rightarrow \frac{(x-0)^2}{25}+\frac{(y-5)^2}{21}=1\\\\\Rightarrow \frac{x^2}{25}+\frac{(y-5)^2}{21}=1$

Therefore, an equation for the ellipse that satisfies the given conditions is  $\frac{x^2}{25}+\frac{(y-5)^2}{21}=1$

Learn more about the ellipse here:

brainly.com/question/28168673

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