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

Let the tangent at ($\alpha , \beta$) be $4x\alpha -y\beta =36$ As ( 0,3 ) lies on the tangent, so we have

$-3\beta =36\Rightarrow \beta =-12$ 

Now $4{\alpha }^2-{\beta }^2=36$$4{\alpha }^2-{12}^2=36$

$\Rightarrow 4{\alpha }^2=180\Rightarrow {\alpha }^2=45\Rightarrow \alpha =\pm 3\sqrt{5}$

Thus the points $P$ and $Q$ are $P(3\sqrt{5},-12),Q(-3\sqrt{5},-12)$

The area of the $∆TQP$ is given by

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

$=\mid \frac{1}{2}\left[-3\left(6\sqrt{5}\right)-36\sqrt{5}-36\sqrt{5}\left|\right|=\frac{1}{2}90\sqrt{5}=45\sqrt{5}\right.$