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JEE Advanced 2015Paper 1

Exam: JEE Advanced
Year: 2015
Paper: Paper 1
Subject: Mathematics
Chapter: Parabola

JEE Advanced 2015 · Paper 1 · Mathematics · Parabola

Let P and Q be distinct points on the parabola y ^ 2 = 2x such that...

Let

P

and

Q

be distinct points on the parabola

{y}^{2} = 2x

such that a circle with

PQ

as diameter passes through the vertex

O

of the parabola. If

P

lies in the first quadrant and the area of the triangle

\delta OPQ

is

3\sqrt{2},

then which of the following is (are) the coordinates of

P

?

Let P and Q be distinct points on the parabola y ^ 2 = 2x such that a circle with PQ as diameter passes through the

Options

A $\left(4, 2\sqrt{2}\right)$
B $\left(9, 3\sqrt{2}\right)$
C $\left(\dfrac{1}{4}, \dfrac{1}{\sqrt{2}}\right)$
D $\left(1, \sqrt{2}\right)$

Correct answer

A. $\left(4, 2\sqrt{2}\right)$

Step-by-step solution

Let P and Q be distinct points on the parabola y ^ 2 = 2x such that a circle with PQ as diameter passes through the

$∆ = 3\sqrt{2}$

$\dfrac{1}{2}\left|$ \begin{cases}0 & 0 \\ {t}_{1}^{2}2 & {t}_{1} \\ {t}_{2}^{2}2 & {t}_{2} \\ 0 & 0\end{cases} $\right|$

$\dfrac{1}{2}\left(\dfrac{{t}_{1}^{2}{t}_{2}}{2} - \dfrac{{t}_{2}^{2}{t}_{1}}{2}\right) = \pm 3\sqrt{2}$

$\dfrac{1}{4}{t}_{1}{t}_{2}\left({t}_{1} - {t}_{2}\right) = \pm 3\sqrt{2}$

${t}_{1}{t}_{2}\left({t}_{1} - {t}_{2}\right) = \pm 12\sqrt{2}$

${t}_{1} - {t}_{2} = \pm 3\sqrt{2}$

$⇒ {t}_{1} - {t}_{2} = 3\sqrt{2}$ (as ${t}_{1} > {t}_{2}$ )

${t}_{1} + {t}_{2} = \sqrt{18 - 16} = \pm \sqrt{2}$

$2{t}_{1} = 3\sqrt{2} + \sqrt{2}$

${t}_{1} = 2\sqrt{2}$

$2{t}_{1} = 2\sqrt{2}$

${t}_{1} = \sqrt{2}$

$ $P\left(4, 2\sqrt{2}\right)$ and $P\left(1, \sqrt{2}\right)

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