The point P ( - 2 6 , 3 ) lies on the hyperbola x 2 a 2 - y 2 b 2 = 1 having eccentricity 5 2 . If the tangent and normal at P to the hyperbola intersect its conjugate axis at the points Q and R respectively, then Q R is equal to:

Step-by-step solution

As point P ( - 2 6 , 3 ) lies on hyperbola. 24 a 2 - 3 b 2 = 1       . . . 1 Given, eccentricity e = 5 2 ⇒ e = a 2 + b 2 a 2 = 5 2 ⇒ 4 a 2 + 4 b 2 = 5 a 2 ⇒ a 2 = 4 b 2       . . . 2 On solving equations 1 and 2 , we get ⇒ 6 b 2 - 3 b 2 = 1 ⇒ b 2 = 3   &   a 2 = 12 Hyperbola x 2 12 - y 2 3 = 1 2 x 12 - 2 y · y ' 3 = 0 y ' | ( - 2 6 , 3 ) = - 1 2 Equation of tangent is ( y - 3 ) = - 1 2 ( x + 2 6 ) at x = 0 ,   y = - 3 , &#16