Question

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 point $C, D$. Then the area of the quadrilateral $ACBD$ is equal to

• A. $\frac{9}{2}$
• B. $\frac{3 \sqrt{17}}{2}$
• C. $\frac{3 \sqrt{17}}{4}$
• D. $9$

Answer 1

Answer: (B)

$ ( x -1)^2+( y -2)^2+( x +2)^2+( y -1)^2=14$
$\Rightarrow x ^2+ y ^2+ x -3 y -2=0$
Put $ x =0$
$ \Rightarrow y ^2-3 y -2=0$
$ \Rightarrow y =\frac{3 \pm \sqrt{17}}{2}$
Put $ y =0 $
$ \Rightarrow x ^2+ x -2=0 $
$ ( x +2)( x -1)=0$
$\therefore A (-2,0), B (1,0), C \left(0, \frac{3+\sqrt{17}}{2}\right), D \left(0, \frac{3-\sqrt{17}}{2}\right) $
Area $=\frac{1}{2} \cdot 3 \cdot \sqrt{17}=\frac{3 \sqrt{17}}{2}$