How do you write an equation of an ellipse in standard form given Vertices: (-9,18), (-9,2), Foci (-9,14), (-9,2)?

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How do you write an equation of an ellipse in standard form given Vertices: (-9,18), (-9,2), Foci (-9,14), (-9,2)?

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Answer 1 · 2017-01-28T22:25:50

The foci are $(h, k - \sqrt{a^{2} - b^{2}}) and (h, k + \sqrt{a^{2} - b^{2}})$ $(h,k-sqrt(a^2-b^2)) and (h,k+sqrt(a^2-b^2))$ for an ellipse with a vertical major axis. The vertices are $(h, k - a) and (h, k + a)$ $(h,k-a) and (h,k+a)$
The equation is $\frac{{(y - k)}^{2}}{a^{2}} + \frac{{(x - h)}^{2}}{b^{2}} = 1 [1]$ $(y-k)^2/a^2+(x-h)^2/b^2=1" [1]"$

Explanation:

Use the given vertices to write 3 equations:

$h = - 9 [2]$ $h = -9" [2]"$
$k - a = 2 [3]$ $k-a=2" [3]"$
$k + a = 18 [4]$ $k+a=18" [4]"$

To find the value of k, add equations [3] and [4]:

$2 k = 20$ $2k = 20$

$k = 10$ $k = 10$

To find the value of a, substitute 10 for k into equation [4]:

$10 + a = 18$ $10+a=18$

$a = 8$ $a = 8$

Use $a = 8, k = 10$ $a = 8, k = 10$ and the y coordinate of one of the foci to write an equation where b is the only unknown:

$14 = 10 + \sqrt{8^{2} - b^{2}}$ $14 = 10 + sqrt(8^2 - b^2)$

$4 = \sqrt{8^{2} - b^{2}}$ $4 = sqrt(8^2 - b^2)$

$16 = 8^{2} - b^{2}$ $16 = 8^2 - b^2$

$- 48 = - b^{2}$ $-48 = -b^2$

$b = 4 \sqrt{3}$ $b = 4sqrt(3)$

Substitute, $h = - 9, k = 10, a = 8, and b = 4 \sqrt{3}$ $h = -9,k=10, a = 8, and b = 4sqrt3$ into equation [1]:

$\frac{{(y - 10)}^{2}}{8^{2}} + \frac{{(x - - 9)}^{2}}{{(4 \sqrt{3})}^{2}} = 1 \leftarrow$ $(y-10)^2/8^2+(x- -9)^2/(4sqrt3)^2=1" "larr$ the answer.

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How do you write an equation of an ellipse in standard form given Vertices: (-9,18), (-9,2), Foci (-9,14), (-9,2)?

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