Question

If sum of distances of a point from the origin and lines $x=2$ is $4$, then its locus is

• A. $x^2-12y=36$
• B. $y^2+12x=36$
• C. $y^2-12x=36$
• D. $x^2+12y=36$

Answer 1

The correct option is B $y^2+12x=36$

Solution:- (B) $y^2+12x=36$

Let the point be $(h,k)$.

Distance of the point $(h,k)$ from the origin $(0,0)$-

$d_1=\sqrt{{(h-0)}^2+{(k-0)}^2}=\sqrt{h^2+k^2}$

Distance of the point $(h,k)$ from the line $x-2=0$-

$d_2=\frac{h-2}{\sqrt{1^2}}=h-2$

Given that the sum of distance of the point $(h,k)$ from the origin and the line $x=2$ is $4$.

$\therefore \sqrt{h^2+k^2}+(h-2)=4$

$\Rightarrow \sqrt{h^2+k^2}=4-h+2$

$\Rightarrow \sqrt{h^2+k^2}=6-h$

Squaring both sides, we have

${\left(\sqrt{h^2+k^2}\right)}^2={(6-h)}^2$

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

$\Rightarrow k^2+12h=36$

Replacing $h$ with $x$ and $k$ with $y$, we have

$y^2+12x=36$

Hence the locus of the point is $y^2+12x=36$.