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If the distance from a variable point $\mathrm{P}$ to a fixed point $\mathrm{A}(a, 0)$ is equal to the perpendicular distance from $\mathrm{P}$ to the line $x+y=0$ then the equation of the locus of $\mathrm{P}$ is
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Verified Answer
The correct answer is:
$x^2+y^2-2 x y-4 a x+2 a^2=0$
Given point $\mathrm{A}(a, 0)$ and the line $x+y=0$
Let point $\mathrm{P}$ is $(h, k)$
According to question,
$$
\sqrt{(h-a)^2+(k-0)^2}=\frac{h+k}{\sqrt{2}}
$$
Take square both sides,
$$
\begin{aligned}
& h^2+a^2-2 h a+k^2=\frac{1}{2}\left(h^2+k^2+2 h k\right) \\
& 2 h^2+2 k^2-4 h a+2 a^2=h^2+k^2+2 h k \\
& h^2+k^2-4 h a-2 h k+2 a^2=0 \\
& \text { Replace }(h, k) \rightarrow(x, y) \\
& x^2+y^2-4 x a-2 x y+2 a^2=0
\end{aligned}
$$
Therefore, option (b) is correct.
Let point $\mathrm{P}$ is $(h, k)$
According to question,
$$
\sqrt{(h-a)^2+(k-0)^2}=\frac{h+k}{\sqrt{2}}
$$
Take square both sides,
$$
\begin{aligned}
& h^2+a^2-2 h a+k^2=\frac{1}{2}\left(h^2+k^2+2 h k\right) \\
& 2 h^2+2 k^2-4 h a+2 a^2=h^2+k^2+2 h k \\
& h^2+k^2-4 h a-2 h k+2 a^2=0 \\
& \text { Replace }(h, k) \rightarrow(x, y) \\
& x^2+y^2-4 x a-2 x y+2 a^2=0
\end{aligned}
$$
Therefore, option (b) is correct.
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