QJEEM25Maef1e: JEE Main 2025 Mathematics - Class 11 JEEM25Maef1e Solution: Two parabolas have the same focus (4, 3) and their directrices…

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

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

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

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