Question

11 Tangent and normal are drawn at any point Pon the parabola $y^2=4\sqrt{\sqrt{2}}x$ such that tangent intersect $x$-axis at $Q$ and normal intersect the curve at $R$. Chord $PR$ subtends right angle at vertex of parabla If area of triangle PQR is $\Delta$, then find $\sqrt{\Delta }$.

Answer 1

Answer: 6

$y^2=4\sqrt{\sqrt{2}}x$
$a=\sqrt{\sqrt{2}}$
$t_1{t}_2=-4$
(For $90^{\circ }$ at vertex)
$t_2=-t_1-\frac{2}{t_1}$ (condition for normal to meet again at parabola)

Solving (1) and (2)
$\frac{-4}{t_1}=t_1-\frac{2}{t_1}$
$t_1{}^2=2$
$t_1=\pm \sqrt{2}$
$\text{ Take }{t}_1=\sqrt{2}$
$t_2=-2\sqrt{2}$
$\text{ Points }P(2a,2\sqrt{2}a),Q(-2a,0),R(8a,-4\sqrt{2}a)$
$\Delta (PQR)=\frac{1}{2}\left|\begin{array}{ccc}-2a& 0& 1\\ 2a& 2\sqrt{2}a& 1\\ 8a& -4\sqrt{2}a& 1\end{array}\right|=\frac{1}{2}\left|-12\sqrt{2}{a}^2-24\sqrt{2}{a}^2\right|$
$\frac{36\sqrt{2}{a}^2}{2}=18\sqrt{2}{a}^2=18\times \sqrt{2}\cdot \sqrt{2}=18\times 2$
$\Delta =36\Rightarrow \sqrt{\Delta }=6$