Question

Let the pair of tangents from $P(3,4)$ touch the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ at $A(\alpha ,0)$ and $B(\beta ,\gamma )$. If $G$ is the centroid of $△PAB$ and the area of $△GAB$ is equal to $k$ sq. units, then the value of $5k$ is

Answer 1

$\frac{x^2}{9}+\frac{y^2}{4}=1\cdots \left(1\right)$
Equation of chord $AB$ is $T=0$
$\Rightarrow \frac{3x}{9}+\frac{4y}{4}=1$
$\Rightarrow x+3y=3\cdots \left(2\right)$

On solving equation $\left(1\right)$ and $\left(2\right)$, we get the points of chord of contact as $A\equiv (3,0)$ and $B\equiv (-\frac{9}{5},\frac{8}{5})$

Area of $△GAB,k=\frac{1}{3}\left(\text{area of}△PAB\right)$
$=\frac{1}{3}\times \frac{1}{2}\left|\left|\begin{array}{ccc}3& 4& 1\\ 3& 0& 1\\ -\frac{9}{5}& \frac{8}{5}& 1\end{array}\right|\right|=\frac{16}{5}$
$\therefore 5k=5\times \frac{16}{5}=16$