JEE Advanced 2026 — Paper 1 — MATH-I solution

JEE Advanced2026Paper 1MATH-I

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 $x^{2} + 4 y^{2} = 8$ such that the slope of the tangent to the ellipse at the point R is $- \frac{1}{2}$ . Then the radius of the circle passing through the points P, Q and R is

A
$10$
B
$5$
C
$\frac{5}{2}$
D
$25$

Solution

Finding P: For $y = x^{2}$ , $\frac{d y}{d x} = 2 x = 4 \Rightarrow x = 2$ , so $P = (2, 4)$ .

Finding Q: For $x^{2} + y^{2} = 2$ , implicit differentiation gives $\frac{d y}{d x} = - \frac{x}{y} = - 1 \Rightarrow x = y$ . With $x^{2} + x^{2} = 2$ and the first-quadrant condition, $Q = (1, 1)$ .

Finding R: For $x^{2} + 4 y^{2} = 8$ , $\frac{d y}{d x} = - \frac{x}{4 y} = - \frac{1}{2} \Rightarrow x = 2 y$ . Substituting, $4 y^{2} + 4 y^{2} = 8 \Rightarrow y = 1$ , so $R = (2, 1)$ .

Radius: Observe that $P = (2, 4)$ , $Q = (1, 1)$ , $R = (2, 1)$ . The segment $P R$ is vertical and $QR$ is horizontal, so $∠ P R Q = 9 0^{\circ}$ . Hence $P Q$ is the diameter of the circumscribing circle.

$P Q = (2 - 1)^{2} + (4 - 1)^{2} = 10$ , so the radius is $\frac{10}{2} = \frac{5}{2}$ .

$∴$ The correct option is (C).

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