A point $P(x, y)$ is such that the sum of squares of its distance from $(a, 0)$ and $(-a, 0)$ is $2 b^2$.…

A point $P(x, y)$ is such that the sum of squares of its distance from $(a, 0)$ and $(-a, 0)$ is $2 b^2$. The equation representing the locus of $P$ is
  1. $x^2+y^2=b^2+a^2$
  2. $x^2+y^2=b^2-a^2$
  3. $x^2+y^2=b^2-2 a^2$
  4. $x^2+y^2=b^2+2 a^2$

Solution

Let the point be $(x, y)$. $ \begin{aligned} & \Rightarrow \quad(x-a)^2+(y-0)^2+(x+a)^2+(y-0)^2=2 b^2 \\ & \Rightarrow \quad 2 x^2+2 y^2+2 a^2=2 b^2 \\ & \Rightarrow \quad x^2+y^2=b^2-a^2 . \end{aligned} $

Asked in: AP EAMCET 2022 (07 Jul Shift 2)

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