JEE Main 2025 — 3 April, Morning Shift — Question 37

The given parabolas are symmetric about the line $y=x$.

Tangents at $\mathrm{A} \& \mathrm{~B}$ must be parallel to $\mathrm{y}=\mathrm{x}$ line, so slope of the tangents $=1$

$\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\min \mathrm{A}}=1=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\min \mathrm{B}}$

For point $\mathrm{B}, \quad \mathrm{y}=\mathrm{x}^{2}+2$

$$\frac{d y}{d x}=2 x=1$$

$x=\frac{1}{2} \Rightarrow y=\frac{9}{4}$

$\therefore$ Point $\mathrm{B}=\left(\frac{1}{2}, \frac{9}{4}\right) \Rightarrow$ Point $\mathrm{A}=\left(\frac{9}{4}, \frac{1}{2}\right)$

$\mathrm{AB}=\sqrt{\left(\frac{1}{2}-\frac{9}{4}\right)^{2}+\left(\frac{9}{4}-\frac{1}{2}\right)^{2}}$

$=\sqrt{\frac{98}{16}}=\frac{7 \sqrt{2}}{4}$

Radius $=\frac{7 \sqrt{2}}{8}$