Let A = (a, 0) and B = (-a, 0) be two fixed p | vedclass

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(B) Let the coordinates of point $P$ be $(x, y)$.Given $PA = nPB$,squaring both sides gives $PA^2 = n^2 PB^2$.Substituting the coordinates,we get $(x

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$A$ circle

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(B) Let the coordinates of point $P$ be $(x, y)$.Given $PA = nPB$,squaring both sides gives $PA^2 = n^2 PB^2$.Substituting the coordinates,we get $(x - a)^2 + (y - 0)^2 = n^2 ((x + a)^2 + (y - 0)^2)$.Expanding the terms: $x^2 - 2ax + a^2 + y^2 = n^2 (x^2 + 2ax + a^2 + y^2)$.Rearranging the terms: $x^2(1 - n^2) + y^2(1 - n^2) - 2ax(1 + n^2) + a^2(1 - n^2) = 0$.Since $|n| 
eq 1$,we have $1 - n^2 
eq 0$. Dividing by $(1 - n^2)$,we get $x^2 + y^2 - 2ax \frac{1 + n^2}{1 - n^2} + a^2 = 0$.This is the equation of a circle of the form $x^2 + y^2 + 2gx + 2fy + c = 0$.

Let $A = (a, 0)$ and $B = (-a, 0)$ be two fixed points. For $a \in (-\infty, 0)$ and a point $P$ moving in the plane such that $PA = nPB$ $(n \neq 0)$. If $|n| \neq 1$,then the locus of point $P$ is....

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