Question

If P(m, n) is a point on an ellipse (x²/a²) + (y²/b²) = 1  with foci S and S' and eccentricity e, then area of SPS' is ........

(a) ae√(a² – m²)

(b) ae√(b² – m²)

(c) be√(b² – m²)

(d) be√(a² – m²)

Answer 1

The correct option (d) be√(a² – m²)

Explanation:

Ellipse: (x²/a²) + (y²/b²) = 1

∵ (m, n) lie on it ⇒ (m²/a²) + (n²/b²) = 1

∴   (n²/b²) = 1 – (m²/a²)

∴     n = b√[1 – (m²/a²)]  .......(1)

Area of SPS' = (1/2)n(SS')

= (1/2)(n)(2ae)

= aeb√[1 – (m²/a²)]

= eb√(a² – m²).