question_answer

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 \[\Delta \text{PTQ}\] is: [JEE Main Online 08-04-2018]

A)  \[60\sqrt{3}\]             

B)  \[36\sqrt{5}\]

C)  \[45\sqrt{5}\]             

D)  \[54\sqrt{3}\]

Correct Answer: C

Solution :

                \[\frac{{{x}^{2}}}{9}-\frac{{{y}^{2}}}{36}=1\] Equation of \[P{{Q}_{(T=0)}}\to 0-\frac{y}{12}=1\] \[\because \text{P  }\!\!\And\!\!\text{  Q are (}\pm \text{3}\sqrt{5},-12)\] \[\therefore \text{ Area =}\frac{1}{2}6\sqrt{5}\times 15=45\sqrt{5}\]