JEE Main 2025 (Online) 29th January Morning ShiftQuestion 1
Two parabolas have the same focus $(4,3)$ and their directrices are the $x$-axis and the $y$-axis, respectively. If these parabolas intersect at the points $A$ and $B$, then $(AB)^2$ is equal to :
Parabola
Solution
Concept. A common point is at equal distance from the focus and from both directrices, so $x^2=y^2$. Formulas used. (1) $SP=PM$: distance to the $x$-axis is $|y|$, to the $y$-axis is $|x|$.(2) $(x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2$.Step 1 - Equations. $(x-4)^2+(y-3)^2=y^2$ and $(x-4)^2+(y-3)^2=x^2$. Step 2 - Subtract. $x^2=y^2\Rightarrow y=x$ or $y=-x$. Step 3 - Case y = x. $(x-4)^2+(x-3)^2=x^2\Rightarrow x^2-14x+25=0$, discriminant $196-100=96>0$: two points. Step 4 - Case y = -x. $(x-4)^2+(x+3)^2=x^2\Rightarrow x^2-2x+25=0$, discriminant $<0$: no point. Step 5 - Use (2). $(x_1-x_2)^2=196-100=96$. On $y=x$, $(y_1-y_2)^2=(x_1-x_2)^2$, so $$(AB)^2=2\cdot96=192.$$ Answer: C