Question

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

• A. \(\frac{9}{2}\)
• B. \(\frac{3\sqrt17}{2}\)
• C. \(\frac{3\sqrt17}{4}\)
• D. 9

Answer 1

The Correct Option is B

Let point P be (h, k)
\((ℎ–1)^2+(k–2)^2+(ℎ+2)^2+(k–1)^2=14\)
\(2ℎ^2+2k^2+2ℎ–6k–4=0\)
Locus of point P : x² + y² + x – 3y – 2 = 0
Intersection with x-axis,
x² + x – 2 = 0
x = –2, 1
Intersection with y-axis,
y² – 3y – 2 = 0
\(y=\frac{3±\sqrt17}{2}\)
Area of the quadrilateral ACBD is \(=\frac{1}{2}(|x1|+|x2|)(|y1|+|y2|)\)
\(=\frac{1}{2}×3×\sqrt17=\frac{3\sqrt17}{2}\)
So, the correct option is (B): \(\frac{3\sqrt17}{2}\)