JEE Main 2025 — Parabola Question with Solution

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

$$\begin{array}{rl}& \therefore (\mathrm{x}-4{)}^2+(\mathrm{y}-3{)}^2={\mathrm{x}}^2\text{..... (1)}\\ & \&(\mathrm{x}-4{)}^2+(\mathrm{y}-3{)}^2={\mathrm{y}}^2\text{..... (2)}\end{array}$$

Here $A\left(x_1,y_1\right)\&B\left(x_2,y_2\right)$ will satisfy the equation

Also from equations (1) & (2), we get $x=y$ .......(3)

Put $x=y$ in equation (1)

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

$$\begin{array}{rl}& x_1+x_2=14\\ & x_1{x}_2=25\\ & \therefore A{B}^2={\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2\\ & =2{\left(x_1-x_2\right)}^2\\ & =2\left[{\left(x_1+x_2\right)}^2-4x_1{x}_2\right]\\ & =192\end{array}$$