The length of the latus-rectum of the ellipse, whose foci are (2,5) and (2,-3) and eccentricity is 4/5, is

Solution

Maths sir · 4 steps

The distance between the foci is 2ae. That gives a, then b², then the latus rectum.

1. Use the foci. Both foci lie on the line x = 2, so the major axis is vertical. The distance between them is 5 - (-3) = 8, and this distance equals 2ae.
2. Find a. 2ae = 8 gives ae = 4. With e = 4/5, this gives a = 4 × 5/4 = 5.
3. Find b squared. b² = a²(1 - e²) = 25(1 - 16/25) = 9, so b = 3. The vertical major axis does not change this formula.

   Latus rectum = 2b²/a = (2 × 9)/5 = 18/5

4. So. The length of the latus rectum is 18/5, which is option (b).

The idea to keep. For any ellipse, the distance between the foci is 2ae and the latus rectum is 2b²/a, with a always the semi-major axis. A vertical major axis changes the picture, not the formulas.

Answer from the NTA final answer key. Solution by PrepWiser.