Question

The area of triangle formed by the tangents from point $(3,2)$ to hyperbola $x^2-9y^2=9$ and the chord of contact w.r.t. point $(3,2)$ is :

• A. $8$
• B. $4$
• C. $6$
• D. $16$

Answer 1

Answer: (A)

$8$

Hyperbola is $x^2-9y^2=9$ or $\frac{x^2}{9}-\frac{y^2}{1}=1$

Equation of tangent is $y=mx\pm \sqrt{a^2{m}^2-b^2}$ ..... (i)

It passes through $(3,2)$

$\Rightarrow 2=3m\pm \sqrt{9m^2-1}$

or $4+9m^2-12m=9m^2-1$

$\Rightarrow m_1=\frac{5}{12}$ and $m_2=\infty$

Equation of tangent (i) for $m_1=\frac{5}{12}$

$y=\frac{5}{12}x\pm \sqrt{9(\frac{5}{12}{)}^2-1}$

or $y=\frac{5}{12}x\pm \frac{3}{4}$

On taking (-)ve sign, point $P(3,2)$ does not satisfy the equation of tangent therefore rejecting (-)ve sign.

Hence equation of tangent is $y=\frac{5x}{12}+\frac{3}{4}$ .....(ii)

Now, equation of tangent (i) for $m_2=\infty$ is $x\pm 3=0$

rejecting $(+)$ sign (since taking $(+)$ sign point $P(3,2)$ does not satisfy this equation.)

Hence, equation of tangent is $x-3=0$ ....(iii)

Now equation of chord of contact w.r.t. point $P(3,2)$ is $T=0$

or $3x-18y=9$

or $x-6y=3$ .....(iv)

Solving (ii) and (iv); $x=-5,y=-\frac{4}{3}$

Solving (iii) and (iv); $x=3,y=0$

Now vertices of triangle are $(3,2),(3,0),(-5,-\frac{4}{3})$

$\therefore Area=\frac{1}{2}\left|\begin{array}{ccc}3& 2& 1\\ 3& 0& 1\\ -5& -\frac{4}{3}& 1\end{array}\right|$

$=\frac{1}{2}|4-16-4|$

$=8sq.units$