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Q.
The locus of a point whose sum of the distances from the origin and the line $x=2$ is 4 units, is
Conic Sections
Solution:
Let the coordinates of the point be $(h, k)$.
Distance of the point from origin
$=\sqrt{(h-0)^{2}+(k-0)^{2}}=\sqrt{h^{2}+k^{2}}$
Distance of the point from the line $x-2=0$ is $h-2$
According to the given condition,
$\sqrt{h^{2}+k^{2}}+h-2=4 $
$\Rightarrow \sqrt{h^{2}+k^{2}}=6-h$
Squaring both sides, we have,
$h^{2}+k^{2}=36+h^{2}-12 h$ or $k^{2}=-12(h-3)$
$\therefore $ The path of the point is $y^{2}=-12(x-3)$.