JEE Main & Advanced
JEE Main Online Paper (Held On 08 April 2018)
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 \[P{{Q}_{(T=0)}}\to 0-\frac{y}{12}=1\] \[\because \text{P }\!\!\And\!\!\text{ Q are (}\pm \text{3}\sqrt{5},-12)\] \[\therefore \text{ Area =}\frac{1}{2}6\sqrt{5}\times 15=45\sqrt{5}\]