Tangents are drawn to the hyperbola 4 x ^ 2 - y ^ 2 = 36 at the...

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 sq. units)$ of \delta PTQ is: Tangents are drawn to the hyperbola 4 x ^ 2 - y ^ 2 = 36 at the points P and Q . If these tangents intersect at the

Options

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

Correct answer

B. $45\sqrt{5}$

Step-by-step solution

Given hyperbola

4{x}^{2} - {y}^{2} = 36

⇒ \dfrac{{x}^{2}}{9} - \dfrac{{y}^{2}}{36} = 1

Equation of chord of contact w.r.t. point

T ( 0, 3 )

\dfrac{x{x}_{1}}{9} - \dfrac{y{y}_{1}}{36} = 1

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

⇒ y = - 12

This intersect the hyperbola

\dfrac{{x}^{2}}{9} - \dfrac{144}{36} = 1

x = \pm 3\sqrt{5}

P\left(3\sqrt{5}, - 12\right)andQ\left(- 3\sqrt{5}, - 12\right)

Hence, area of

∆ = \dfrac{1}{2} \times 6\sqrt{5} \times 15 = 45\sqrt{5}

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