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

$\quad y^2=4 \sqrt{\sqrt{2}} x$
$a=\sqrt{\sqrt{2}}$
$t _1 t _2=-4$
(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 (2 a , 2 \sqrt{2} a ), \quad Q (-2 a , 0), \quad R (8 a ,-4 \sqrt{2} a ) $
$ \Delta( PQR )=\frac{1}{2}\begin{vmatrix}-2 a & 0 & 1 \\ 2 a & 2 \sqrt{2} a & 1 \\ 8 a & -4 \sqrt{2} a & 1\end{vmatrix}=\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 $