Question

A point P moves so that its distance from the point $(a,0)$ is always equal to its distance from the line $x + a = 0$. Find the locus of the point.
A.${y^2} = 4ax$
B.${x^2} = 4ay$
C.${y^2} + 4ax = 0$
D.${x^2} + 4ay = 0$

Answer 1

Hint: First suppose the coordinate of the point. Then obtain the distance between the point and the given point. Write the mathematical expression for the stated problem. Equate the obtained distance and the given line equation. Square both sides of the obtained equation and calculate to obtain the required locus.

Formula Used:
The distance between the points $(a,b),(c,d)$ is
$\sqrt {{{\left( {c - a} \right)}^2} + {{\left( {d - b} \right)}^2}} $ .
${a^2} - {b^2} = (a + b)(a - b)$

Complete step by step solution:
Suppose the coordinate of the point is $A(x,y)$ .
The distance of AP is,
$\sqrt {{{\left( {a - x} \right)}^2} + {{(0 - y)}^2}} $
$=\sqrt {{{\left( {a - x} \right)}^2} + {y^2}} $
It is given that,
$\sqrt {{{\left( {a - x} \right)}^2} + {y^2}} = a + x$
Square both sides of the equation,
${\left( {\sqrt {{{\left( {a - x} \right)}^2} + {y^2}} } \right)^2} = {\left( {a + x} \right)^2}$
${\left( {a - x} \right)^2} + {y^2} = {\left( {a + x} \right)^2}$
${y^2} = {\left( {a + x} \right)^2} - {\left( {a - x} \right)^2}$
$ = (a + x + a - x)(a + x - a + x)$
${y^2} = 2a.2x$
${y^2} = 4ax$

Option ‘A’ is correct

Note: The distance of the point $P\left( {h,k} \right)$ from the line $x + a = 0$ is $\left| {\dfrac{{h + a}}{{\sqrt {{1^2}} }}} \right|$.
The distance of the point $P\left( {h,k} \right)$ from $\left( {a,0} \right)$ is $\sqrt {{{\left( {h - a} \right)}^2} + {{\left( {k - 0} \right)}^2}} $
According to question,
$\sqrt {{{\left( {h - a} \right)}^2} + {{\left( {k - 0} \right)}^2}} = \left| {\dfrac{{h + a}}{{\sqrt {{1^2}} }}} \right|$
Square both sides
$ \Rightarrow {\left( {h - a} \right)^2} + {\left( {k - 0} \right)^2} = {\left( {h + a} \right)^2}$
$ \Rightarrow {h^2} - 2ha + {a^2} + {k^2} = {h^2} + 2ha + {a^2}$
$ \Rightarrow {k^2} = 4ha$
The equation of locus of the point is ${y^2} = 4ax$.