Question

The locus of a point 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,

$h^2+k^2=36+h^2-12h\Rightarrow k^2=-12(h-3)$

$\therefore$ The path of the point is $y^2=-12(x-3)$