Question
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 $$T\left( {0,\,3} \right)$$ then the area (in square units) of $$\Delta PTQ$$ is :
A.
$$54\sqrt 3 $$
B.
$$60\sqrt 3 $$
C.
$$36\sqrt 5 $$
D.
$$45\sqrt 5 $$
Answer :
$$45\sqrt 5 $$
Solution :
Here equation of hyperbola is $$\frac{{{x^2}}}{9} - \frac{{{y^2}}}{{36}} = 1$$
Now, $$PQ$$ is the chord of content
$$\therefore $$ Equation of $$PQ$$ is :
$$\eqalign{ & \frac{{x\left( 0 \right)}}{9} - \frac{{y\left( 3 \right)}}{{36}} = 1 \cr & \Rightarrow y = - 12 \cr} $$
$$\eqalign{ & \therefore {\text{Area of }}\Delta PQT = \frac{1}{2} \times TR \times PQ \cr & \because P \equiv \left( {3\sqrt 5 ,\, - 12} \right) \cr & \therefore TR = 3 + 12 = 15 \cr & \therefore {\text{Area of }}\Delta PQT = \frac{1}{2} \times 15 \times 6\sqrt 5 = 45\sqrt 5 \,\,{\text{sq}}{\text{. units}} \cr} $$
Here equation of hyperbola is $$\frac{{{x^2}}}{9} - \frac{{{y^2}}}{{36}} = 1$$
Now, $$PQ$$ is the chord of content
$$\therefore $$ Equation of $$PQ$$ is :
$$\eqalign{ & \frac{{x\left( 0 \right)}}{9} - \frac{{y\left( 3 \right)}}{{36}} = 1 \cr & \Rightarrow y = - 12 \cr} $$
$$\eqalign{ & \therefore {\text{Area of }}\Delta PQT = \frac{1}{2} \times TR \times PQ \cr & \because P \equiv \left( {3\sqrt 5 ,\, - 12} \right) \cr & \therefore TR = 3 + 12 = 15 \cr & \therefore {\text{Area of }}\Delta PQT = \frac{1}{2} \times 15 \times 6\sqrt 5 = 45\sqrt 5 \,\,{\text{sq}}{\text{. units}} \cr} $$