Question

Let P be a point on the ellipse (x²/a²) + (y²/b²) = 1 of eccentricity e. If A, A' are the vertices and S, S' are the foci of an ellipse, then area of APA' : area of PSS' = .......

(a) e 

(b) e² 

(c) e³

(d) (1/e)

Answer 1

The correct option (d) (1/e)

Explanation:

Ellipse is (x²/a²) + (y²/b²) = 1

area of APA' = (1/2)(AA')(b sinθ)

area of PSS' = (1/2)(SS')(b sin θ)

∴ Ration = (AA'/SS')

= [(distance between vertices)/(distance between foci)]

= (2a/2ae)

= (1/e).