If two points P and Q lie on the hyperbola x 2 a 2 - y 2 b 2 = 1 ( a < b ) , whose centre C be such that C P is perpendicular to C Q , then the value of 1 C P 2 + 1 C Q 2 is

Step-by-step solution

Let C P = r 1 be inclined to transverse axis at an angle θ such that P r 1 cos ⁡ θ , r 1 sin ⁡ θ lies on the hyperbola. Therefore, r 1 2 cos 2 ⁡ θ a 2 - sin 2 ⁡ θ b 2 = 1 Replacing θ b y 90 o + θ , we get, r 2 2 sin 2 ⁡ θ a 2 - cos 2 ⁡ θ b 2 = 1 (where, C Q = r 2 ) ⇒ 1 r 1 2 + 1 r 2 2 = cos 2 ⁡ θ a 2 - sin 2 ⁡ θ b 2 + sin 2 ⁡ θ a 2 - cos 2 ⁡ θ b 2 ⇒ 1 r 1 2 + 1 r 2 2 = cos 2 ⁡ θ 1 a 2 - 1 b 2 + sin 2 ⁡ θ 1 a 2 - 1 b 2 ⇒ 1 r 1 2 + 1 r 2 2 = 1 a 2 - 1 b 2 ⇒ 1 C P 2 + 1 C Q 2 = 1 a 2 - 1 b 2