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If the sum of the squares of the distances of the point $(x, y)$ from the points $(a, 0)$ and $(-a, 0)$ is $2 b^{2}$, then which one of the following is correct?
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Verified Answer
The correct answer is:
$x^{2}+a^{2}=b^{2}-y^{2}$
Let $\mathrm{P}(x, y)$ be a point and $\mathrm{A}=(a, 0), \mathrm{B}=(-a, 0)$. Now, $\mathrm{PA}^{2}=(x-a)^{2}+y^{2}$
$\mathrm{PB}^{2}=(x+a)^{2}+y^{2}$
Since the sum of the distances of the point $\mathrm{P}(x, y)$ from the points $\mathrm{A}(a, 0)$ and $\mathrm{B}(-a, 0)$ is $2 b^{2}$. $\therefore \quad \mathrm{PA}^{2}+\mathrm{PB}^{2}=2 b^{2}$
$(x-a)^{2}+(y)^{2}+(x+a)^{2}+(y)^{2}=2 b^{2}$
$\Rightarrow x^{2}+a^{2}-2 a x+y^{2}+x^{2}+a^{2}+2 a x+y^{2}=2 b^{2}$
$\Rightarrow x^{2}+a^{2}+y^{2}=b^{2}$
$\Rightarrow x^{2}+a^{2}=b^{2}-y^{2}$
$\mathrm{PB}^{2}=(x+a)^{2}+y^{2}$
Since the sum of the distances of the point $\mathrm{P}(x, y)$ from the points $\mathrm{A}(a, 0)$ and $\mathrm{B}(-a, 0)$ is $2 b^{2}$. $\therefore \quad \mathrm{PA}^{2}+\mathrm{PB}^{2}=2 b^{2}$
$(x-a)^{2}+(y)^{2}+(x+a)^{2}+(y)^{2}=2 b^{2}$
$\Rightarrow x^{2}+a^{2}-2 a x+y^{2}+x^{2}+a^{2}+2 a x+y^{2}=2 b^{2}$
$\Rightarrow x^{2}+a^{2}+y^{2}=b^{2}$
$\Rightarrow x^{2}+a^{2}=b^{2}-y^{2}$
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