Tangents are drawn to the hyperbola 4 x 2 - y 2 = 36 at the points P and Q . If these tangents intersect at the point T 0 , 3 then the area (in sq. units) of Δ P T Q is:

Step-by-step solution

Given hyperbola 4 x 2 - y 2 = 36 ⇒ x 2 9 - y 2 36 = 1 Equation of chord of contact w.r.t. point T ( 0 ,   3 ) : x x 1 9 - y y 1 36 = 1 x . 0 9 - y . 3 36 = 1 ⇒   y =   - 12 This intersect the hyperbola x 2 9 - 144 36 = 1   at x = ± 3 5 P 3 5 ,   - 12   a n d   Q - 3 5 ,   - 12 Hence, area of ∆ = 1 2 × 6 5 × 15 = 45 5