JEE Main 2025 — 29 January, Morning Shift — Question 47

Let intersection points of these two parabolas are $\mathrm{A}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \& \mathrm{~B}\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)$

$\because$ equation of parabola I and II are given below

$\therefore(\mathrm{x}-4)^{2}+(\mathrm{y}-3)^{2}=\mathrm{x}^{2}$

$\&(x-4)^{2}+(y-3)^{2}=y^{2}$

Here $\mathrm{A}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \& B\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)$

will satisfy with equation Also from equations (1) & (2), we get $x=y$

Put $x=y$ in equation (1)

We get $x^{2}-14 x+25=0$

$\mathrm{x}_{1}+\mathrm{x}_{2}=14$||$\mathrm{x}_{1} \mathrm{x}_{2}=25$

$\therefore \mathrm{AB}^{2}=\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)^{2}+\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)^{2}$

$=2\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)^{2}$

$=2\left[\left(\mathrm{x}_{1}+\mathrm{x}_{2}\right)^{2}-4 \mathrm{x}_{1}\mathrm{x}_{2}\right]$

$=192$