JEE Advanced 2026 · Paper 2 · Mathematics Q2

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MathematicsSingle correct+3 / -1Question 2 of 54

Let $T$ be the tangent to the parabola $y^2=16x$ at the point $(64,32)$ . Let $L$ be the tangent to the same parabola at another point $(x_1,y_1)$ on the parabola. If $L$ and $T$ are perpendicular to each other, then the distance between the point $(x_1,y_1)$ and the focus of the parabola, is

\dfrac{15}4

\dfrac{17}4

Answer and worked solution

Official answer: C

Here $a=4$ ; points of the parabola are $(4t^2,8t)$ and the tangent at $t$ has slope $\frac1t$ .

$(64,32)$ is $t=4$ . Perpendicular tangents need $t_1t_2=-1$ , so $t_2=-\frac14$ , giving $(x_1,y_1)=\left(\frac14,-2\right)$ .

The focal distance is $x_1+a=\frac14+4=\dfrac{17}4$ .

Topics:Parabola Tangents and normals to conics

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JEE Advanced 2026 · Paper 2, Mathematics question 2