Question

The equation of the locus of a point whose distance from $\left( {a,0} \right)$ is equal to its distance from $y$-axis is
A. ${y^2} - 2ax = {a^2}$
B. ${y^2} - 2ax + {a^2} = 0$
C. ${y^2} + 2ax + {a^2} = 0$
D. ${y^2} + 2ax = {a^2}$

Answer 1

Hint: Locus is a path obtained by movement of a point satisfying one or more than one conditions. Here the condition is the distance of the point from $\left( {a,0} \right)$ and from $y$-axis are equal. Let the coordinates of the point, whose locus is to be determined, be $\left( {h,k} \right)$. Then find the distances of this point from $\left( {a,0} \right)$ and from $y$-axis and make an equation by equating the distances. After simplification of this equation, an equation in terms of $h$ and $k$ will be obtained. Replacing $h$ by $x$ and $k$ by $y$, the equation of the locus will be obtained.

Formula Used:
Distance between two points having coordinates $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $ units.
Distance of the point $\left( {a,b} \right)$ from $y$-axis is $a$ units.

Complete step by step solution:
Let the required point be $P$ and its coordinates are $\left( {h,k} \right)$ and the given point is $A\left( {a,0} \right)$.
Find the distance between the points $P$ and $A$.
Here ${x_1} = h,{y_1} = k,{x_2} = a,{y_2} = 0$
Using the distance formula, we get $\bar A\bar P = \sqrt {{{\left( {a - h} \right)}^2} + {{\left( {0 - k} \right)}^2}} $
The distance of the point $\left( {h,k} \right)$ is $h$ units.
According to the given condition, these two distances are equal.
So, $\sqrt {{{\left( {a - h} \right)}^2} + {{\left( {0 - k} \right)}^2}} = h$
Simplify this equation.
Square both sides of the equation.
$ \Rightarrow {\left( {a - h} \right)^2} + {\left( {0 - k} \right)^2} = {h^2}$
Use the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$
$ \Rightarrow {a^2} - 2ah + {h^2} + {k^2} = {h^2}$
Cancel out the term ${h^2}$ from both sides
$ \Rightarrow {a^2} - 2ah + {k^2} = 0$
Arrange the terms
$ \Rightarrow {k^2} - 2ah + {a^2} = 0$
Replace $h$ by $x$ and $k$ by $y$
$ \Rightarrow {y^2} - 2ax + {a^2} = 0$
This is the required equation of the locus of the point $P\left( {h,k} \right)$.

Option ‘B’ is correct

Note: The distance of a point from $x$-axis is equal to its ordinate and the distance of the point from $y$-axis is equal to its abscissa. You should not take the coordinates of the moving point as $\left( {x,y} \right)$ because the point is at a fixed position at a particular time. At last you have to replace the abscissa of the moving point by $x$ and the ordinate by $y$.