Question

The locus of a point whose sum of the distances from the origin and the line $x = 2$ is 4 units is
A.${y^2} = - 12\left( {x - 3} \right)$
B.${y^2} = 12\left( {x - 3} \right)$
C.${x^2} = 12\left( {y - 3} \right)$
D.${x^2} = - 12\left( {y - 3} \right)$

Answer 1

Here, we will assume that the coordinates of the point be $\left( {h,k} \right)$ and compare the formula to find the distance of the point from origin is $l = \sqrt {{{\left( {{x_1} - 0} \right)}^2} + {{\left( {{y_1} - 0} \right)}^2}}$ from the above figure, where $\left( {{x_1},{y_1}} \right)$ is the point to find the ${x_1}$ and ${y_1}$, Then we know that the distance of the point from the line $x - 2 = 0$ is $h - 2$ and find the sum of the distance formula to the point from the origin and the distance from the line. After simplifying, we will replace $x$ for $h$ and $y$ for $k$ in the above equation to find the path of the point.

Complete step-by-step answer:
We are given that the locus of a point whose sum of the distances from the origin and the line $x = 2$ is 4 units.

Let us assume that the coordinates of the point be $\left( {h,k} \right)$.
We know that the formula to find the distance of the point from origin is $l = \sqrt {{{\left( {{x_1} - 0} \right)}^2} + {{\left( {{y_1} - 0} \right)}^2}}$ from the above figure, where $\left( {{x_1},{y_1}} \right)$ is the point.
Finding the value of ${x_1}$ and ${y_1}$, we get
${x_1} = h$
${y_1} = k$
Substituting the values of ${x_1}$ and ${y_1}$ in the above formula of distance, we get

$$\Rightarrow \sqrt {{{\left( {h - 0} \right)}^2} + {{\left( {k - 0} \right)}^2}} \\\ \Rightarrow \sqrt {{h^2} + {k^2}} \\\$$

Now, we know that the distance of the point from the line $x - 2 = 0$ is $h - 2$.
So, according to the given condition we have to find the sum of the distance formula to the point from the origin and the distance from the line.
$\Rightarrow \sqrt {{h^2} + {k^2}} + h - 2 = 4$
Subtracting the above equation by $h - 2$ on both sides, we get

$$\Rightarrow \sqrt {{h^2} + {k^2}} + h - 2 - \left( {h - 2} \right) = 4 - \left( {h - 2} \right) \\\ \Rightarrow \sqrt {{h^2} + {k^2}} + h - 2 - h + 2 = 4 - h + 2 \\\ \Rightarrow \sqrt {{h^2} + {k^2}} = 6 - h \\\$$

Squaring both sides of the above equation, we have

$$\Rightarrow {h^2} + {k^2} = {\left( {6 - h} \right)^2} \\\ \Rightarrow {h^2} + {k^2} = {h^2} + 36 - 12h \\\$$

Subtracting the above equation by ${h^2}$ on both sides, we get

$$\Rightarrow {h^2} + {k^2} - {h^2} = {h^2} + 36 - 12h - {h^2} \\\ \Rightarrow {k^2} = 36 - 12h \\\ \Rightarrow {k^2} = - 12\left( {x - 3} \right) \\\$$

Replacing $x$ for $h$ and $y$ for $k$ in the above equation to find the path of the point, we get
$\Rightarrow {y^2} = - 12\left( {x - 3} \right)$
Hence, option A is correct.

Note: In solving these types of questions, the key concept is to remember that the distance of any point from $x$–axis and $y$–axis would be $y$ and $x$ respectively. Always remember to recall the distance formula to get to the required result.