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

If the normal at the point P ( θ ) to the ellipse x 2 14 + y 2 5 = 1 , intersects it again at the point Q ( 2 θ ) show that cosθ = - 2 3

Given equation of ellipse is x 2 14 + y 2 5 = 1 Here, a 2 = 14 , b 2 = 5 Let P ( θ ) = a cos θ , b sin θ Now, equation of normal at point P ( θ ) is given by 14 x cosθ - 5 y sinθ = 9 . . . . ( 1 ) Now, let Q ( 2 θ ) = a cos 2 θ , b sin 2 θ As equation ( 1 ) passes through Q ( 2 θ ) , ⇒ 14 . 14 cos 2 θ cosθ - 5 . 5 sin 2 θ sinθ = 9 ⇒ 14 cos 2 θ cosθ - 10 cosθ = 9 ⇒ 14 cos 2 θ - 10 cos 2 θ = 9 cosθ ⇒ 14 2 cos 2 θ - 1 - 10 cos 2 θ = 9 cosθ ⇒ 18 cos 2 θ - 9 cosθ - 14 = 0 ⇒ 18 cos 2 θ - 21 cosθ + 12 cosθ - 14 = 0 ⇒ 3 cosθ ( 6 cosθ - 7 ) + 2 ( 6 cosθ - 7 ) = 0 ⇒ cosθ = - 2 3