JEE Advanced 2026 · Paper 1 · Mathematics Q2

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MathematicsSingle correct+3 / -1Question 2 of 48

Let $P$ be the point on the parabola $y=x^2$ such that the slope of the tangent to the parabola at the point $P$ is $4$ . Let $Q$ be the point in the first quadrant lying on the circle $x^2+y^2=2$ such that the slope of the tangent to the circle at the point $Q$ is $-1$ . Let $R$ be the point in the first quadrant lying on the ellipse $x^2+4y^2=8$ such that the slope of the tangent to the ellipse at the point $R$ is $-\frac12$ . Then the radius of the circle passing through the points $P,Q$ and $R$ is

\sqrt{10}

\sqrt5

\sqrt{\dfrac52}

2\sqrt5

Answer and worked solution

Official answer: C

The three points.

On $y=x^2$ : $2x=4$ , so $P=(2,4)$ .
On $x^2+y^2=2$ : slope $-x/y=-1$ , so $x=y=1$ and $Q=(1,1)$ .
On $x^2+4y^2=8$ : slope $-\dfrac{x}{4y}=-\dfrac12$ , so $x=2y$ ; then $8y^2=8$ , giving $R=(2,1)$ .

Circle. $Q$ and $R$ both have $y=1$ , so the centre has $x=\frac32$ . $P$ and $R$ both have $x=2$ , so the centre has $y=\frac52$ .

Centre $\left(\frac32,\frac52\right)$ , and
$r^2=\left(\tfrac12\right)^2+\left(\tfrac32\right)^2=\tfrac52\implies r=\sqrt{\tfrac52}.$

Topics:Equation of a circle Tangents and normals to conics Tangents and normals

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JEE Advanced 2026 · Paper 1, Mathematics question 2