JEE MAIN - Mathematics (2022 - 26th July Morning Shift - No. 10)
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 $$\mathrm{P}$$, which intersects the $$x$$-axis at the points $$\mathrm{A}$$, $$\mathrm{B}$$ and the $$y$$-axis at the points C, D. Then the area of the quadrilateral ACBD is equal to :
$${9 \over 2}$$
$${{3\sqrt {17} } \over 2}$$
$${{3\sqrt {17} } \over 4}$$
9
Explanation
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
$$ = {1 \over 2}(|{x_1}| + |{x_2}|)(|{y_1}| + |{y_2}|)$$
$$ = {1 \over 2} \times 3 \times \sqrt {17} = {{3\sqrt {17} } \over 2}$$
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