QJEEAdv15Ma7db6: JEE Advanced 2015 Mathematics - Class 11 JEEAdv15Ma7db6 Solution: Let and be distinct points on the parabola such that a circle…

Let $P\left( {{{t_1^2} \over 2},{t_1}} \right)$ and $Q\left( {{{t_2^2} \over 2},{t_2}} \right)$ be two distinct points on the parabola ${y^2} = 2x$ .

The circle with PQ as diameter passes through the vertex O(0, 0) of the parabola.

Clearly, PO $\bot$ OQ

So, slope of PO $\times$ slope of OQ = $-$ 1

or, ${{{t_1} - 0} \over {{{t_1^2} \over 2} - 0}} \times {{{t_2} - 0} \over {{{t_2^2} \over 2} - 0}} = - 1$

or, ${2 \over {{t_1}}} \times {2 \over {{t_2}}} = - 1$

or, ${t_1}{t_2} = - 4$

By question, area of $\Delta OPQ = 3\sqrt 2$

or, $$\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ \frac{t_1^2}{2}& t_1& 1\\ \frac{t_2^2}{2}& t_2& 1\end{array}\right|=3\sqrt{2}$$

or, ${1 \over 2}\left| {{{t_1^2{t_2}} \over 2} - {{{t_1}t_2^2} \over 2}} \right| = 3\sqrt 2$

or, $\left| {{t_1}{t_2}} \right|\left| {{t_1} - {t_2}} \right| = 12\sqrt 2$

or, $\left| {{t_1} + {4 \over {{t_1}}}} \right| = 3\sqrt 2$ [ $\because$ ${t_1}{t_2} = - 4$ ]

or, ${t_1} + {4 \over {{t_1}}} = 3\sqrt 2$ [ $\because$ P lies in first quadrant]

or, $t_1^2 - 3\sqrt 2 {t_1} + 4 = 0$

or, ${t_1} = {{3\sqrt 2 \pm \sqrt {18 - 4 \times 1 \times 4} } \over 2}$

$= {{3\sqrt 2 \pm \sqrt 2 } \over 2}$

$= 2\sqrt 2 ,\sqrt 2$

$\therefore$ coordinates of $P = (4,2\sqrt 2 )$ or $(1,\sqrt 2 )$ .

Therefore, (A) and (D) are correct options.