JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Circle

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

Answer: (B)

Solution

Let

P = (x, y)

. The condition gives:

(x - 1)^{2} + (y - 2)^{2} + (x + 2)^{2} + (y - 1)^{2} = 14

Expanding:

2 x^{2} + 2 y^{2} + 2 x - 6 y + 10 = 14

, simplifying to

x^{2} + y^{2} + x - 3 y - 2 = 0

.

For x-intercepts (

y = 0

x^{2} + x - 2 = 0

, giving

x = 1

x = - 2

. So

A = (- 2, 0)

B = (1, 0)

.

For y-intercepts (

x = 0

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

, giving

y = \frac{3 \pm 17}{2}

. So

C = (0, \frac{3 + 17}{2})

D = (0, \frac{3 - 17}{2})

.

The quadrilateral ACBD has diagonals AB (length 3) and CD (length

17

) perpendicular to each other.

Area

= \frac{1}{2} \times 3 \times 17 = \frac{3 17}{2}

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Single correct option