JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Parabola

For the parabola with focus $(4,3)$ and directrix the $x$-axis, a point $(x,y)$ satisfies


For the one with directrix the $y$-axis,


Subtracting, $x^2-y^2=0$, so $x=y$ or $x=-y$.

Substituting $x=-y$ into (1) gives $x^2-2x+25=0$, whose discriminant is negative, so there is no real solution there.

Substituting $x=y$ into (1) gives

with roots $x_{1,2}=7\pm 2\sqrt{6}$ (so $x_1+x_2=14$, $x_1{x}_2=25$). The two intersection points are $(x_1,x_1)$ and $(x_2,x_2)$, so