Question

Tangents to the parabola $y^2=4x$ at $P$ and $Q$ meet at $T(x_1,0)$ and normals at $P$ and $Q$ meet at $R(x_2,0).$ If $x_2$ is the length of the latus rectum of the ellipse $3x^2+8y^2=48,$ then the area of the quadrilateral $PTQR$ is

Answer 1

$3x^2+8y^2=48\Rightarrow \frac{x^2}{16}+\frac{y^2}{6}=1$
$a^2=16,b^2=6$
Length of latus rectum =$\frac{2b^2}{a}=3$
Here $t^2+2=3\Rightarrow t^2=1\Rightarrow t=1$(for point $P$ )
Area of the quadrilateral
$=2\times \text{Area of}\Delta PTR=2\times \frac{1}{2}\times PR\times TP$
$=2\sqrt{2}\times 2\sqrt{2}$
$=8$