If C is the centre of a hyperbola x 2 a 2 - y 2 b 2 = 1 , S , S ' its foci and P a point on it. Prove that S P · S ' P = C P 2 - a 2 + b 2 .

Let, the point P on the hyperbola x 2 a 2 - y 2 b 2 = 1 is P a sec θ , b tan θ . We know that the focal distance of a point x 1 , y 1 on the hyperbola are e x 1 ± a . Hence, we have S P · S ' P = a e sec θ - a × a e secθ + a ⇒ S P · S ' P = a 2 e 2 sec 2 θ - a 2 ⇒ S P · S ' P = a 2 1 + b 2 a 2 sec 2 θ - a 2 ⇒ S P · S ' P = a 2 + b 2 sec 2 θ - a 2 ⇒ S P · S ' P = a 2 sec 2 θ + b 2 sec 2 θ - a 2 Using 1 + tan 2 θ = sec 2 θ ⇒ S P · S ' P = a 2 sec 2 θ + b 2 1 + tan 2 θ - a 2 ⇒ S P · S ' P = a 2 sec 2 θ + b 2 tan 2 θ + b 2 - a 2 . . . 1 Now, the distance of the point P a sec θ , b tan θ from the centre C 0 , 0 is C P = a 2 sec 2 θ + b 2 tan 2 θ , ⇒ S P · S ' P = C P 2 + b 2 - a 2 .

Tangents are drawn to the hyperbola 3 x 2 - 2 y 2 = 25 from the point 0 , 5 2 . Find their equations.

Given hyperbola is 3 x 2 - 2 y 2 = 25 . ⇒ x 2 25 3 - y 2 25 2 = 1 . . . 1 Tangent to hyperbola x 2 a 2 - y 2 b 2 = 1 having the slope m is y = m x + a 2 m 2 - b 2 . Thus, the equation of tangent to the hyperbola 1 is y = m x + 25 3 m 2 - 25 2 . . . 2 It passes through the point 0 , 5 2 . Therefore 5 2 = 25 m 2 3 - 25 2 ⇒ 25 4 = 25 m 2 3 - 25 2 ⇒ 25 m 2 3 = 25 4 + 25 2 ⇒ 100 m 2 = 75 + 150 ⇒ m 2 = 9 4 ⇒ m = ± 3 2 . So, putting m in the equation 2 , the equation of tangents are y = ± 3 2 x + 25 3 · 9 4 - 25 2 ⇒ y = ± 3 2 x + 75 4 - 50 4 ⇒ y = ± 3 2 x + 5 2 .

If the normal at a point P to the hyperbola x 2 a 2 - y 2 b 2 = 1 meets the x axis at G , show that S G = e . S P , S being the focus of the hyperbola.

We know that the foci of the hyperbola x 2 a 2 - y 2 b 2 = 1 are S = a e , 0 and S ' = - a e , 0 . We also know that the equation of normal of the hyperbola x 2 a 2 - y 2 b 2 = 1 at P a sec θ , b tan θ is a x cos θ + b y cotθ = a 2 + b 2 . Given that the normal of the hyperbola x 2 a 2 - y 2 b 2 = 1 at P a sec θ , b tan θ meets the x -axis at G . ⇒ G = a 2 + b 2 a cos θ , 0 Now, S G S P = a e - a 2 + b 2 a cos θ 2 + 0 - 0 2 a e - a sec θ 2 + 0 - b tan θ 2 ⇒ S G S P = a e - 1 + b 2 a 2 sec θ a e - sec θ 2 + b 2 a 2 tan 2 θ ⇒ S G S P = e 1 - e sec θ e - sec θ 2 + e 2 - 1 tan 2 θ ⇒ S G S P = e 1 - e sec θ e 2 + sec 2 θ - 2 e sec θ + e 2 tan 2 θ - tan 2 θ ⇒ S G S P = e 1 - e sec θ e 2 sec 2 θ - 2 e sec θ + 1 ⇒ S G S P = e 1 - e sec θ 1 - e sec θ 2 ⇒ S G = e · S P