Let a and b be positive real numbers such that a > 1 and b < a . Let P be a point in the first quadrant that lies on the hyperbola x 2 a 2 - y 2 b 2 = 1 . Suppose the tangent to the hyperbola at P passes through the point 1 , 0 , and suppose the normal to the hyperbola at P cuts off equal intercepts on the coordinate axes. Let Δ denote the area of the triangle formed by the tangent at P , the normal at P

Step-by-step solution

Tangent at P x sec θ a - y tan θ b = 1 Passes through 1 , 0 sec θ = a Now slope of AP = 1 b sec θ a tan θ = 1 ⇒ b = tan θ Now b 2 = a 2 e 2 - 1 tan 2 θ = sec 2 θ e 2 - 1 ⇒ e 2 - 1 = sin 2 θ e 2 - 1 ∈ 0 , 1 ∵ θ ∈ 0 , π 2 e 2 ∈ 1 , 2 ⇒ 1 e 2 Now A 1 , 0 P a s e c θ , b tan θ = s e c 2 θ , tan 2 θ A P = tan 4 θ + tan 4 θ ⇒ A P = 2 tan 2 θ ∵ ∆ A P B is isosceles-right angled triangle, So, area = 1 2 A P 2 = 1 2 × 2 tan 4 θ = b 4 .