The coordinates of the points A and B are (ak | vedclass

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(D) Let the coordinates of $P$ be $(x, y)$.Given $PA = kPB$,squaring both sides gives $PA^2 = k^2 PB^2$.Using the distance formula: $(x - ak)^2 + y^2

Vedclass · Accepted Answer

${x^2} + {y^2} - {a^2} = 0$

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(D) Let the coordinates of $P$ be $(x, y)$.Given $PA = kPB$,squaring both sides gives $PA^2 = k^2 PB^2$.Using the distance formula: $(x - ak)^2 + y^2 = k^2 \left[ (x - \frac{a}{k})^2 + y^2 ight]$.Expanding the terms: $x^2 - 2akx + a^2k^2 + y^2 = k^2 \left[ x^2 - 2x(\frac{a}{k}) + \frac{a^2}{k^2} + y^2 ight]$.$x^2 - 2akx + a^2k^2 + y^2 = k^2x^2 - 2akx + a^2 + k^2y^2$.Rearranging the terms: $x^2(1 - k^2) + y^2(1 - k^2) = a^2 - a^2k^2$.$(1 - k^2)(x^2 + y^2) = a^2(1 - k^2)$.Since $k = \pm 1$,$k^2 = 1$,so $1 - k^2 = 0$. However,for the locus to be defined as a standard equation independent of $k$ when $k 
eq 1$,we simplify the expression: $x^2 + y^2 = a^2$.Thus,the equation is $x^2 + y^2 - a^2 = 0$.

The coordinates of the points $A$ and $B$ are $(ak, 0)$ and $(\frac{a}{k}, 0)$,where $k = \pm 1$. If a point $P(x, y)$ moves such that $PA = kPB$,then the equation to the locus of $P$ is:

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