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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 $A C B D$ is equal to

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$\frac{9}{2}$

$\frac{3 \sqrt{17}}{2}$

$\frac{3 \sqrt{17}}{4}$

answer is B.

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Detailed Solution

According to question, ${(x - 1)}^{2} + {(y - 2)}^{2} + {(x + 2)}^{2} + {(y - 1)}^{2} = 14$

$\Rightarrow x^{2} + y^{2} + x - 3 y - 2 = 0$ …………. (i)

Put $x = 0$ . In eq. (i) $\Rightarrow y^{2} - 3 y - 2 = 0$

Apply quadratic formula, $\Rightarrow y = \frac{3 \pm \sqrt{17}}{2}$

Put $y = 0,$ in eq. (i). $\Rightarrow x^{2} + x - 2 = 0$

$x = - 2, 1 .$

Then, point $A (- 2, 0), B (1, 0) C (0, \frac{3 + \sqrt{17}}{2}) & D (0, \frac{3 - \sqrt{17}}{2})$

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