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 :

Solution

Maths sir · 5 steps

Get the circle, read off where it cuts each axis, and use half the product of the perpendicular diagonals.

1. Write the locus. Let P = (x, y). Then (x-1)² + (y-2)² + (x+2)² + (y-1)² = 14. Expanding gives 2x² + 2y² + 2x - 6y + 10 = 14, that is, x² + y² + x - 3y - 2 = 0, a circle.
2. Cut the x-axis. Put y = 0: x² + x - 2 = 0, so x = 1 or x = -2. So A and B are (1, 0) and (-2, 0), and AB = 3.
3. Cut the y-axis. Put x = 0: y² - 3y - 2 = 0, so y = (3 ± √17)/2. The two roots differ by √17, so CD = √17. The product of the roots is -2, so C and D lie on opposite sides of the origin.
4. Diagonals are perpendicular. In ACBD the diagonals are AB, along the x-axis, and CD, along the y-axis. Both pass through the origin and they are perpendicular, so the area is half the product of the diagonals.

   Area = 1/2 × AB × CD = 1/2 × 3 × √17

5. So. The area of ACBD is 3√17/2, which is option (b).

The idea to keep. When a curve cuts both axes on either side of the origin, the four points form a quadrilateral with perpendicular diagonals. Its area is half the product of the two intercept lengths.

Answer from the NTA final answer key. Solution by PrepWiser.