JEE Main 2022 (Online) 26th July Morning ShiftQuestion 1
A point $P$ moves so that the sum of squares of its distances from the points $(1,2)$ and $(-2,1)$ is $14$. Let $f(x,y)=0$ be the locus of $P$, which intersects the $x$-axis at the points $A$, $B$ and the $y$-axis at the points $C$, $D$. Then the area of the quadrilateral $ACBD$ is equal to :
Circle
Solution
$(x-1)^2+(y-2)^2+(x+2)^2+(y-1)^2=14\Rightarrow x^2+y^2+x-3y-2=0$. $x$-axis: $x^2+x-2=0\Rightarrow x=1,-2$, $AB=3$. $y$-axis: $y^2-3y-2=0$, $CD=|y_1-y_2|=\sqrt{9+8}=\sqrt{17}$. Diagonals are perpendicular: area $=\frac12\cdot3\cdot\sqrt{17}=\frac{3\sqrt{17}}{2}$. Option (B).