+2 votes
in JEE by kratos

If (x/2a) + [(y√3)/2b] = 1 touches the ellipse (x2/a2) + (y2/b2) = 1, then its eccentric angle θ of the contact point is .......

(a) 0°

(b) 60°

(c) 45°

(d) 90°

1 Answer

+3 votes
by kratos
 
Best answer

The correct option (b) 60°

Explanation:

For given ellipse, tangent at point of contact is

[(x cosθ)/a] + [(ysinθ)/b] = 1 .....(1)

also given tangent is (x/a) ∙ (1/2) + (y/b) ∙ (√3/2) = 1 .....(2)

as (1) & (2) are identical

⇒ cos θ = (1/2) and sin θ = (√3/2)

⇒ θ = (π/3) = 60°.

...