Area Ques 15
- Let $P$ and $Q$ be distinct points on the parabola $y^2=2 x$ such that a circle with $P Q$ as diameter passes through the vertex $O$ of the parabola. If $P$ lies in the first quadrant and the area of $\triangle O P Q$ is $3 \sqrt{2}$, then which of the following is/are the coordinates of $P$ ?
(2015 Adv.)
(a) $(4,2 \sqrt{2})$
(b) $(9,3 \sqrt{2})$
(c) $\left(\frac{1}{4}, \frac{1}{\sqrt{2}}\right)$
(d) $(1, \sqrt{2})$
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Answer:
Correct Answer: 15.(a,d)
Solution: (a,d) Since, $\angle P O Q=90^{\circ}$
$\Rightarrow \quad \frac{t_1-0}{\frac{t_1^2}{2}-0} \cdot \frac{t_2-0}{\frac{t_2^2}{2}-0}=-1 \Rightarrow t_1 t_2=-4$
$\because \quad \operatorname{ar}(\triangle O P Q)=3 \sqrt{2}$
$\therefore \quad \frac{1}{2}\left|\begin{array}{ccc}0 & 0 & 1 \\ t_1^2 / 2 & t_1 & 1 \\ t_2^2 / 2 & t_2 & 1\end{array}\right|= \pm 3 \sqrt{2} \Rightarrow \frac{1}{2}\left(\frac{t_1^2 t_2}{2}-\frac{t_1 t_2^2}{2}\right)= \pm 3 \sqrt{2}$
$\Rightarrow \quad \frac{1}{4}\left(-4 t_1+4 t_2\right)= \pm 3 \sqrt{2} \Rightarrow t_1+\frac{4}{t_1}=3 \sqrt{2} \quad \left[\because t_1>0\right.$ for $\left.P\right]$
$\Rightarrow \quad t_1^2-3 \sqrt{2} t_1+4=0 \Rightarrow\left(t_1-2 \sqrt{2}\right)\left(t_1-\sqrt{2}\right)=0$
$\Rightarrow \quad t_1=\sqrt{2}\quad $ or $ \quad 2 \sqrt{2}$
$\therefore \quad P(1, \sqrt{2})\quad $ or $\quad P(4,2 \sqrt{2})$
BETA
