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The sum of the squares of the distances of a moving point from 2 fixed points $A(a, 0)$ and $B(-a, 0)$ is equal to a constant $2 c^2$, then the equation of its locus is
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Verified Answer
The correct answer is:
$x^2+y^2=c^2-a^2$
Let the point be $P(x, y)$.
Then,
$$
\begin{aligned}
& B P=\sqrt{(x+a)^2+(y-0)^2} \\
& \text { and } A P=\sqrt{(x-a)^2+(y-0)^2}
\end{aligned}
$$
Given that,
$$
(B P)^2+(A P)^2=2 c^2
$$
i.e. $(x+a)^2+y^2+(x-a)^2+y^2=2 c^2$
$$
\begin{aligned}
& \Rightarrow x^2+a^2+2 a x+2 y^2+x^2+a^2-2 a x=2 c^2 \\
& \Rightarrow 2 x^2+2 y^2+2 a^2=2 c^2 \\
& \quad \text { or } x^2+y^2=c^2-a^2
\end{aligned}
$$
Desired locus is $x^2+y^2=c^2-a^2$
Then,
$$
\begin{aligned}
& B P=\sqrt{(x+a)^2+(y-0)^2} \\
& \text { and } A P=\sqrt{(x-a)^2+(y-0)^2}
\end{aligned}
$$
Given that,
$$
(B P)^2+(A P)^2=2 c^2
$$
i.e. $(x+a)^2+y^2+(x-a)^2+y^2=2 c^2$
$$
\begin{aligned}
& \Rightarrow x^2+a^2+2 a x+2 y^2+x^2+a^2-2 a x=2 c^2 \\
& \Rightarrow 2 x^2+2 y^2+2 a^2=2 c^2 \\
& \quad \text { or } x^2+y^2=c^2-a^2
\end{aligned}
$$
Desired locus is $x^2+y^2=c^2-a^2$
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