Question:

A point P(x, y) is such that the sum of squares of its distances from (a, 0) and (--a, 0) is \(2b^2\). The equation representing the locus of P is

Show Hint


Distance formula between \((x_1, y_1)\) and \((x_2, y_2)\) is \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\).
Square of the distance is \((x_2-x_1)^2 + (y_2-y_1)^2\).
Set up the equation based on the given condition and simplify to find the locus.
The equation \(x^2+y^2=r^2\) represents a circle centered at origin with radius \(r\).
Updated On: May 26, 2025
  • \( x^2+y^2 = b^2+a^2 \)
  • \( x^2+y^2 = b^2-a^2 \)
  • \( x^2+y^2 = b^2-2a^2 \)
  • \( x^2+y^2 = b^2+2a^2 \)
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

Let the two given points be \(A=(a,0)\) and \(B=(-a,0)\). Let P be \((x,y)\). Distance PA = \(\sqrt{(x-a)^2 + (y-0)^2} = \sqrt{(x-a)^2 + y^2}\). Square of distance PA: \(PA^2 = (x-a)^2 + y^2 = x^2 - 2ax + a^2 + y^2\). Distance PB = \(\sqrt{(x-(-a))^2 + (y-0)^2} = \sqrt{(x+a)^2 + y^2}\). Square of distance PB: \(PB^2 = (x+a)^2 + y^2 = x^2 + 2ax + a^2 + y^2\). Given that the sum of squares of distances is \(2b^2\): \(PA^2 + PB^2 = 2b^2\). \((x^2 - 2ax + a^2 + y^2) + (x^2 + 2ax + a^2 + y^2) = 2b^2\). Combine terms: \(2x^2 + 2y^2 + 2a^2 = 2b^2\). Divide by 2: \(x^2 + y^2 + a^2 = b^2\). Rearrange to get the locus of P: \(x^2 + y^2 = b^2 - a^2\). This represents a circle centered at the origin \((0,0)\) with radius \(\sqrt{b^2-a^2}\), provided \(b^2>a^2\). If \(b^2=a^2\), it's a point (origin). If \(b^2<a^2\), no real locus. This matches option (b). \[ \boxed{x^2+y^2 = b^2-a^2} \]
Was this answer helpful?
0
0