Find the condition that the line x cos α + y sin α = p may be a tangent to the ellipse x 2 a 2 + y 2 b 2 = 1 . Also find the point of contact.

Given line is x cos α + y sin α = p . . . . . . . 1 and curve is x 2 a 2 + y 2 b 2 = 1 . . . . . . . . . 2 ⇒ b 2 x 2 + a 2 y 2 = a 2 b 2 Now, differentiating equation 2 with respect to x we get ⇒ d y d x = - x b 2 y a 2 From equation 1 , ⇒ y = - x c o t α + p sin α Thus, slope of line is - c o t α So, the given equation of line will be tangent to equation ( 2 ) , if ⇒ x a 2 cos α = y b 2 sin α = k ⇒ x = k a 2 cos α , y = b 2 k sin α So, the line x cos α + y sin α = p will touch the curve x 2 a 2 + y 2 b 2 = 1 at a point ( k a 2 cos α , b 2 k sin α ) From equation 1 ⇒ a 2 cos 2 α + b 2 sin 2 α 2 = p 2 k 2 . . . . . . 3 From equation 2 ⇒ a 2 cos 2 α + b 2 sin 2 α = 1 k 2 . . . . . . . 4 Divide equation 3 by 4 , we get ⇒ a 2 cos 2 α + b 2 sin 2 α = p 2 which is the required condition. And the point of contact will be ( a 2 cos α p , b 2 sin α p ) .

If the eccentric angles of the ends of a focal chord of the ellipse x 2 a 2 + y 2 b 2 = 1 be θ and ψ , show that tan θ 2 tan ψ 2 = e - 1 e + 1 .

Equation of line joining points θ , ψ is x a cos θ + ψ 2 + y b sin θ + ψ 2 = cos θ - ψ 2 If it is a focal chord, then it passes through focus a e , 0 , so e cos θ + ψ 2 = cos θ - ψ 2 ⇒ cos θ - ψ 2 cos θ + ψ 2 = e ⇒ cos θ - ψ 2 - cos θ + ψ 2 cos θ + ψ 2 + cos θ - ψ 2 = e - 1 e + 1 ⇒ 2 sin θ 2 sin ψ 2 2 cos θ 2 cos ψ 2 = e - 1 e + 1 ⇒ tan θ 2 tan ψ 2 = e - 1 e + 1 Now using - a e , 0 , we get tan θ 2 tan ψ 2 = e + 1 e - 1