JEE Main 2022 · 26 Jul · Shift 1 · Mathematics Q12

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MathematicsSingle correct+4 / -1Question 12 of 90

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 :

\dfrac92

\dfrac{3\sqrt{17}}2

\dfrac{3\sqrt{17}}4

Answer and worked solution

Official answer: B

Locus $x^2+y^2+x-3y-2=0$ . x-intercepts: $x^2+x-2=0$ , $AB=3$ . y-intercepts: $y^2-3y-2=0$ , $CD=\sqrt{17}$ .

Area $\frac12\cdot3\sqrt{17}$ (diagonals on the axes).

(B)

Topics:Equation of a circle Locus

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JEE Main 2022 · 26 Jul · Shift 1, Mathematics question 12