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

• A

  $36\sqrt{5}$
• Correct answer
  B

  $45\sqrt{5}$
• C

  $54\sqrt{3}$
• D

  $60\sqrt{3}$

Solution

Given hyperbola $4x^2-y^2=36$

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

Equation of chord of contact w.r.t. point $T(0, 3)$ : $\frac{x{x}_1}{9}-\frac{y{y}_1}{36}=1$

$\frac{x.0}{9}-\frac{y.3}{36}=1$

$\Rightarrow y= -12$

This intersect the hyperbola $\frac{x^2}{9}-\frac{144}{36}=1$

at $x=\pm 3\sqrt{5}$



$P\left(3\sqrt{5}, -12\right) \mathrm{a}\mathrm{n}\mathrm{d} Q\left(-3\sqrt{5}, -12\right)$

Hence, area of $∆=\frac{1}{2}\times 6\sqrt{5}\times 15=45\sqrt{5}$

Hence, the correct answer is option: 'B'.