Question

Derive the equation of locus of points equidistant from point A(1, 0) and the line x=-1

Answer 1

Let $(x,\ y)$ be the point equidistant from the point A(1, 0) and the line $x+1=0.$

The distance between the two points is, by distance formula,

$\longrightarrow d_1=\sqrt{(x-1)^2+y^2}$

The distance of the point $(x,\ y)$ from the line $x+1=0$ is, (use the formula to find perpendicular distance of a point from a line)

$\longrightarrow d_2=\dfrac{|x+1|}{\sqrt{1^2+0^2}}$

$\longrightarrow d_2=|x+1|$

According to the question,

$\longrightarrow d_1=d_2$

$\longrightarrow\sqrt{(x-1)^2+y^2}=|x+1|$

$\longrightarrow(x-1)^2+y^2=(x+1)^2$

$\longrightarrow y^2=(x+1)^2-(x-1)^2$

$\longrightarrow\underline{\underline{y^2=4x}}$

This is the locus of the point $(x,\ y).$