If the straight line l x + m y = n is a normal to the hyperbola x 2 a 2 - y 2 b 2 = 1 , then show that a 2 l 2 - b 2 m 2 = a 2 + b 2 2 n 2 .

Given equation of the hyperbola is x 2 a 2 - y 2 b 2 = 1 Equation of normal to the hyperbola is a x sec θ + b y tan θ = a 2 + b 2 ⇒ a x cos θ + b y cot θ = a 2 + b 2 Given that straight line l x + m y = n is normal to the given hyperbola. ∴ l = a cos θ , m = b cot θ , n = a 2 + b 2 ⇒ a cos θ l = b cot θ m = a 2 + b 2 n ⇒ sec θ = n a l a 2 + b 2 , tan θ = n b m a 2 + b 2 We know that sec 2 θ - tan 2 θ = 1 ⇒ n 2 a 2 l 2 a 2 + b 2 2 - n 2 b 2 m 2 a 2 + b 2 2 = 1 ⇒ a 2 l 2 - b 2 m 2 = a 2 + b 2 2 n 2 Hence, proved.

If the straight line l x + m y = n is a normal to the curve x = a cos ϕ and y = b sin ϕ , 0 ≤ ϕ ≤ 2 π then prove that a 2 l 2 + b 2 m 2 = a 2 - b 2 2 n 2

Given curve is ellipse. Given that line l x + m y = n is a normal to the curve ellipse at a cos ϕ , b sin ϕ . Now, equation of normal at a cos ϕ , b sin ϕ is a x cos ϕ - b y sin ϕ = a 2 - b 2 Comparing this with given line, we get a l cos ϕ = - b m sin ϕ = a 2 - b 2 n ⇒ cos ϕ = a n l a 2 - b 2 ...(1) ⇒ sin ϕ = - b n m a 2 - b 2 ..(2) Squaring and adding (1), (2), we get a 2 n 2 l 2 a 2 - b 2 2 + b 2 n 2 m 2 a 2 - b 2 2 = 1 ∵ sin 2 ϕ + cos 2 ϕ = 1 ⇒ a 2 l 2 + b 2 m = a 2 - b 2 2 n 2 Hence, proved.