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Determine the equation of the ellipse with foci (15,-2) and (3,-2), and co-vertices (9,6) and (9,-10).​

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Answer: We know that the center of the ellipse is the midpoint of the line segment joining the foci. So, the center is ((15+3)/2, (-2-2)/2) = (9, -2).

The distance between the foci is 2c, where c is the distance between the center and each focus. So, 2c = 15 - 3 = 12, which means c = 6.

The distance between the center and each co-vertex is b. So, b = 10 - (-2) = 12.

The semi-major axis is a, which is the distance from the center to a vertex. Since we have the co-vertices, we can use the Pythagorean theorem to find a:

a^2 = b^2 - c^2

a^2 = 12^2 - 6^2

a^2 = 108

a = sqrt(108) = 6*sqrt(3)

Therefore, the equation of the ellipse is:

(x - 9)^2 / (6*sqrt(3))^2 + (y + 2)^2 / 12^2 = 1

Simplifying, we get:

(x - 9)^2 / 72 + (y + 2)^2 / 144 = 1

So, the equation of the ellipse is (x - 9)^2 / 72 + (y + 2)^2 / 144 = 1.

Enjoy!!!!!!!!!!!!!!!!!!!!!!

User Richard Torcato
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