AP EAMCET · Maths · Straight Lines
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
- A \(x^2+y^2=b^2+a^2\)
- B \(x^2+y^2=b^2-a^2\)
- C \(x^2+y^2=b^2-2 a^2\)
- D \(x^2+y^2=b^2+2 a^2\)
Answer & Solution
Correct Answer
(B) \(x^2+y^2=b^2-a^2\)
Step-by-step Solution
Detailed explanation
Let the point be \((x, y)\). \[ \begin{aligned} & \Rightarrow \quad(x-a)^2+(y-0)^2+(x+a)^2+(y-0)^2=2 b^2 \\ & \Rightarrow \quad 2 x^2+2 y^2+2 a^2=2 b^2 \\ & \Rightarrow \quad x^2+y^2=b^2-a^2 . \end{aligned} \]
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