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:

• A. \(x^2+y^2=b^2+a^2\)
• B. \(x^2+y^2=b^2-a^2\)
• C. \(x^2+y^2=b^2-2a^2\)
• D. \(x^2+y^2=b^2+2a^2\)

Hint:

For locus problems, first write the given geometric condition using the distance formula, then simplify algebraically to get the required equation.

Answer 1

The Correct Option is B

Step 1: Set up the distance condition.
Let P(x, y). Distance from (a, 0) is \(\sqrt{(x-a)^2+y^2}\) and from (-a, 0) is \(\sqrt{(x+a)^2+y^2}\). Sum of squares = \(2b^2\).

Step 2: Expand and simplify.
\[(x-a)^2+y^2+(x+a)^2+y^2 = 2b^2\] \[2x^2+2a^2+2y^2 = 2b^2\] \[x^2+y^2 = b^2-a^2\]
\[ \boxed{x^2+y^2 = b^2-a^2} \]