If A = (a, 0) and B = (-a, 0),then the locus | vedclass

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(D) Let the coordinates of point $P$ be $(x, y)$.Given $A = (a, 0)$ and $B = (-a, 0)$.The square of the distance $PA$ is $PA^2 = (x - a)^2 + (y - 0)^2

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a straight line

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(D) Let the coordinates of point $P$ be $(x, y)$.Given $A = (a, 0)$ and $B = (-a, 0)$.The square of the distance $PA$ is $PA^2 = (x - a)^2 + (y - 0)^2 = x^2 - 2ax + a^2 + y^2$.The square of the distance $PB$ is $PB^2 = (x + a)^2 + (y - 0)^2 = x^2 + 2ax + a^2 + y^2$.According to the problem,$PA^2 - PB^2 = a^2$.Substituting the expressions:$(x^2 - 2ax + a^2 + y^2) - (x^2 + 2ax + a^2 + y^2) = a^2$.Simplifying the equation:$-2ax - 2ax = a^2$.$-4ax = a^2$.Since $a 
eq 0$,we divide by $-a$:$4x = -a$,or $x = -\frac{a}{4}$.This is the equation of a straight line parallel to the $y$-axis.

If $A = (a, 0)$ and $B = (-a, 0)$,then the locus of a point $P = (x, y)$ such that $PA^2 - PB^2 = a^2$ is.

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