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…
JEE Main 2022 — Mathematics Coordinate Geometry
2022mcqmedium
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
Official previous-year question
Held on 26 Jul 2022 · Verified 6 Jul 2026.
Options
A
29
B
2317
C
4317
D
9
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Solution
Given,
A point P moves so that the sum of squares of its distances from the points (1,2) and (−2,1) is 14.
So, (x−1)2+(y−2)2+(x+2)2+(y−1)2=14
⇒x2+y2+x−3y−2=0
Put x=0
⇒y2−3y−2=0
⇒y=23±17
Put y=0
⇒x2+x−2=0
(x+2)(x−1)=0
∴A(−2,0),B(1,0),C(0,23+17),D(0,23−17)
Area formed by ACBD will be =21∣−2010−2023+17023−170∣
=21∣−3−17−23−217+23−217+3−17∣
=21×3×17=2317 sq units.
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