Question

The area of triangle (in sq.units) formed by the tangents from point $(3,2)$ to hyperbola $x^2-9y^2=9$ and the chord of contact with respect to the point $(3,2)$ is

Answer 1

Hyperbola is $\frac{x^2}{9}-\frac{y^2}{1}=1$
Equation of tangent is : $y=mx\pm \sqrt{a^2{m}^2-b^2}$
It passes through $(3,2)\Rightarrow 2=3m\pm \sqrt{9m^2-1}$
$\Rightarrow 9m^2+4-12m=9m^2-1$
$\Rightarrow m=\frac{5}{12}$ and it can aso be observed that equation of tangent at vetex is $x=3$,
So, slope of other tangent line is $m=\infty$
$\therefore$ Equation of tangents are :
$x-3=0\cdots \left(1\right)$ and $y=\frac{5}{12}x+\frac{3}{4}\cdots \left(2\right)$
Now, equation of chord of contact is $T=0$ :
$\Rightarrow 3x-18y=9\Rightarrow x-6y=3\cdots \left(3\right)$
Solving $\left(1\right),\left(2\right)$ and $\left(3\right)$, we get point of intersection of three lines as :$(3,2),(3,0),(-5,\frac{-4}{3})$
$\text{Area}=\frac{1}{2}|3(0-\frac{(-4)}{3})+3(\frac{-4}{3}-2)+(-5\left)\right(2-0\left)\right|$
$=\frac{1}{2}|4-4-6-10|$
$=\frac{16}{2}$
$=8$ sq. units

Alternative method:
Hyperbola is $\frac{x^2}{9}-\frac{y^2}{1}=1\Rightarrow x^2-9y^2-9=0\cdots \left(1\right)$
Equation of chord of contact is $T=0$
$\Rightarrow 3x-18y=9\Rightarrow x-6y=3\cdots \left(2\right)$
Using $\left(1\right),\left(2\right)$, we can find the points of intersection of the chord of contact with the hyperbola.
Thus ${(3+6y)}^2-9y^2-9=0$
Thus points are $(3,0),(-5,\frac{-4}{3})$
$\text{Area}=\frac{1}{2}|3(0-\frac{(-4)}{3})+3(\frac{-4}{3}-2)+(-5\left)\right(2-0\left)\right|$
$=\frac{1}{2}|4-4-6-10|$
$=\frac{16}{2}$
$=8$ sq. units