A point P moves so that the sum of squares of its distances from the points (1,2) and (-2,…

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

Accepted Answer

Let point $$P:(h,\,k)$$

$${(h - 1)^2} + {(k - 2)^2} + {(h + 2)^2} + {(k - 1)^2} = 14$$

$2{h^2} + 2{k^2} + 2h - 6k - 4 = 0$

Locus of $P:{x^2} + {y^2} + x - 3y - 2 = 0$

Intersection with x-axis,

${x^2} + x - 2 = 0$

$$ \Rightarrow x =  - 2,\,1$$

Intersection with y-axis,

${y^2} - 3y - 2 = 0$

$$ \Rightarrow y = {{3\, \pm \,\sqrt {17} } \over 2}$$

Area of the quadrilateral ACBD is

$$ = {1 \over 2}(|{x_1}| + |{x_2}|)(|{y_1}| + |{y_2}|)$$

$$ = {1 \over 2} \times 3 \times \sqrt {17}  = {{3\sqrt {17} } \over 2}$$

Solutions · 1