Question

The locus of a point (to the right of $x=2$) whose sum of the distances from the origin and the line $x=2$ is $4$ units, is

• A. $y^2=-12(x-3)$
• B. $y^2=12(x-3)$
• C. $x^2=12(y-3)$
• D. $x^2=-12(y-3)$

Answer 1

The correct option is A $y^2=-12(x-3)$

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
$h^2+k^2=36+h^2-12h\text{or}{k}^2=-12(h-3)$
$\therefore$ The path of the point is $y^2=-12(x-3)$