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)

$(x-1{)}^2+(y-2{)}^2+(x+2{)}^2+(y-1{)}^2=14$
$\Rightarrow x^2+y^2+x-3y-2=0$
Put $x=0$
$\Rightarrow y^2-3y-2=0$
$\Rightarrow y=\frac{3\pm \sqrt{17}}{2}$
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 $=\frac{1}{2}\cdot 3\cdot \sqrt{17}=\frac{3\sqrt{17}}{2}$