SOLUTION: Find the equation given the eccentricity 3/5, distance between directrices 10.

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Question 1167092: Find the equation given the eccentricity 3/5, distance between directrices 10.
Answer by greenestamps(13200) (Show Source):
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The eccentricity is c/a = 3/5. So

Let c = 3x
Let a = 5x

Then gives us b = 4x.

a is the semi-major axis; b is the semi-minor axis; c is the distance from the center to the focus.

The distance from the focus to the end of the major axis is 5x-3x = 2x.

For any point on the ellipse, the eccentricity 3/5 is also the ratio of the distance to the nearest focus and the distance to the nearest directrix. That means the distance from the end of the major axis to the nearest directrix is (5/3)*2x = (10/3)x.

The distance from the center of the ellipse to each directrix is now 5x+(10/3)x = (25/3)x; that makes the distance between the two directrices (50/3)x.

The given distance between the directrices is 10:

And so

a = 5x = 3
b = 4x = 12/5

The equation of the ellipse is