Question

Find the locus of a point if sum of distances of the point from the origin and line $x-2$ is $4$.

• A. $y^2=4(x-1)$
• B. $y^2=-4(x+1)$
• C. $y^2=4(x+1)$
• D. $y^2=-4(x-1)$

Answer 1

Answer: (C)

$y^2=4(x+1)$

Let the point be $(h,k)$

Distance from Origin $(0,0)$ to $(h,k)$ is $\sqrt{(h-0{)}^2+(k-0{)}^2}=\sqrt{h^2+k^2}$ ---------- distance formula $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1)}$

Distance from $x-2$ to $(h,k)$ is $2-h$

Therefore, according to conditions,

$\sqrt{(h2+k2)}+(2-h)=4$

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

$h^2+k^2=(2+h{)}^2$

$h^2+k^2=4+h^2+4h$

$k^2=4(h+1)$

$y^2=4(x+1)$ is the required locus and "Required locus is the parabola"