JEE Advanced 2015 · Paper 1 · Mathematics Q53

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MathematicsMultiple correct+4 / -2Question 53 of 60

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 $3\sqrt2$ , then which of the following is (are) the coordinates of $P$ ?

(4,2\sqrt2)

(9,3\sqrt2)

\left(\dfrac14,\dfrac1{\sqrt2}\right)

(1,\sqrt2)

Answer and worked solution

Official answer: A, D

Parametrise. On $y^2=2x$ , take points $\left(\frac{t^2}2,t\right)$ .

Right angle at O. O lies on the circle with diameter PQ, so $OP\perp OQ$ :
$\frac{t_1^2t_2^2}4+t_1t_2=0\implies t_1t_2=-4.$

Area.
$\text{Area}=\frac14|t_1t_2(t_1-t_2)|=|t_1-t_2|=t_1+\frac4{t_1}=3\sqrt2.$
So $t_1^2-3\sqrt2t_1+4=0$ , giving $t_1=2\sqrt2$ or $\sqrt2$ .

Points. $P=(4,2\sqrt2)$ or $(1,\sqrt2)$ .

(A), (D)

Topics:Parabola Equation of a circle

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JEE Advanced 2015 · Paper 1, Mathematics question 53