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Tangents are drawn to the hyperbola $\mathrm{4 x^2-y^2=36}$ , from the point T(0,3), touching the hyperbola at P and Q, then the area (in sq. units) of $\mathrm{\triangle P T Q}$ is
Option: 1
$36 \sqrt{5}$

Option: 2
$45 \sqrt{5}$

Option: 3
$54 \sqrt{3}$

Option: 4
$60 \sqrt{3}$

Answers (1)

best_answer

Let the tangent at $\mathrm{(\alpha, \beta)}$ be $\mathrm{4 x \alpha-y \beta=36}$

As (0,3) lies on the tangent, so we have

$\mathrm{-3 \beta=36 \Rightarrow \beta=-12}$

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

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

Thus the points P and Q are

$\mathrm{ P(3 \sqrt{5},-12), Q(-3 \sqrt{5},-12) }$

The area of the $\mathrm{\triangle T Q P}$ is given by

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

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

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