Question

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

• A. $60\sqrt{3}$
• B. $36\sqrt{5}$
• C. $45\sqrt{5}$
• D. $54\sqrt{3}$

Answer 1

The correct option is C $45\sqrt{5}$

$\frac{x^2}{9}=\frac{y^2}{36}=1$

eqn of tangent at $(x_1,y_1)$ is given by

$\frac{x{x}_1}{9}-\frac{y{y}_1}{36}=1$

This passes through $(0,3)$

$\therefore 0-\frac{3\left(y_1\right)}{36}=1$

$\Rightarrow y_1=-12$

$\therefore \frac{x_1^2}{9}-\frac{(-12{)}^2}{36}=1$

$\therefore \frac{x_1^2}{9}-4=1$

$\therefore x_1^2=45$

$\therefore x_1=\pm 3\sqrt{5}$

$\therefore P(3\sqrt{5},-12)$ $Q(-3,\sqrt{5},-12)$ $T(o,3)$

$A=\frac{1}{2}\left|\begin{array}{ccc}1& 17& 1\\ 3\sqrt{5}& -3\sqrt{5}& 0\\ -12& -12& 3\end{array}\right|=45\sqrt{5}$