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 need to find the locus of the given point. First, we will use the distance formula to find the distance between the origin and the given point. Then, we will find the distance between the given point and the line $x = 2$. We will use the given information to form an equation in two variables and simplify it using algebraic identities. Finally, we will replace the variables to find the locus of the point.

Formula Used:
We will use the following formulas:
Distance formula: The distance $d$ between two points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is given by the formula $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$.
The distance between a point $\left( {{x_1},{y_1}} \right)$ and a line $ax + by + c = 0$ is given by the formula $d = \dfrac{{\left| {a{x_1} + b{y_1} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}$.

Complete step by step solution:
Let the point be $P\left( {p,q} \right)$.
The origin is the point $\left( {0,0} \right)$.
First, we will use the distance formula to find the distance between the points $\left( {0,0} \right)$ and $\left( {p,q} \right)$.
The distance formula states that the distance $d$ between two points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is given by the formula $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$.
Let ${d_1}$ be the distance between the points $\left( {0,0} \right)$ and $\left( {p,q} \right)$.
Substituting ${x_1} = 0$, ${y_1} = 0$, ${x_2} = p$, and ${y_2} = q$ in the distance formula, we get
$\Rightarrow {d_1} = \sqrt {{{\left( {p - 0} \right)}^2} + {{\left( {q - 0} \right)}^2}}$
Subtracting the like terms in the parentheses, we get
$\Rightarrow {d_1} = \sqrt {{p^2} + {q^2}}$
Now, we will find the distance between the point $\left( {p,q} \right)$ and the line $x = 2$.
Let ${d_2}$ be the distance between the point $\left( {p,q} \right)$ and the line $x = 2$.
The distance between a point $\left( {{x_1},{y_1}} \right)$ and a line $ax + by + c = 0$ is given by the formula $d = \dfrac{{\left| {a{x_1} + b{y_1} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}$.
Rewriting the line $x = 2$, we get
$1 \times x + 0 \times y - 2 = 0$
Comparing the equation $1 \times x + 0 \times y - 2 = 0$ to the equation $ax + by + c = 0$, we get
$a = 1$, $b = 0$, and $c = - 2$
Substituting ${x_1} = p$, ${y_1} = q$, $a = 1$, $b = 0$, and $c = - 2$ in the formula $d = \dfrac{{\left| {a{x_1} + b{y_1} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}$
${d_2} = \dfrac{{\left| {1\left( p \right) + 0\left( q \right) + \left( { - 2} \right)} \right|}}{{\sqrt {{1^2} + {0^2}} }}$
Simplifying the expression, we get
$\Rightarrow$ ${d_2} = \dfrac{{\left| {p + 0 - 2} \right|}}{{\sqrt {1 + 0} }}$
Thus, we get
$\Rightarrow$ $\begin{array}{l}{d_2} = \dfrac{{\left| {p + 0 - 2} \right|}}{{\sqrt {1 + 0} }}\\\\{d_2} = \dfrac{{\left| {p - 2} \right|}}{{\sqrt 1 }}\end{array}$
Simplifying the expression further, we get
$\Rightarrow$ $\begin{array}{l}{d_2} = \left| {p - 2} \right|\\\\{d_2} = p - 2\end{array}$
Now, it is given that the sum of the distance of the point from the origin and the line $x = 2$ is 4 units.
Therefore, we get
$\Rightarrow$ ${d_1} + {d_2} = 4$
Substituting ${d_1} = \sqrt {{p^2} + {q^2}}$ and ${d_2} = p - 2$ in the equation, we get
$\Rightarrow \sqrt {{p^2} + {q^2}} + p - 2 = 4$
Rewriting the equation, we get
$\Rightarrow \sqrt {{p^2} + {q^2}} = 4 + 2 - p$
Adding the terms, we get
$\Rightarrow \sqrt {{p^2} + {q^2}} = 6 - p$
Squaring both sides of the equation, we get
$\Rightarrow {p^2} + {q^2} = {\left( {6 - p} \right)^2}$
Simplifying the expression using algebraic identities, we get
$\Rightarrow {p^2} + {q^2} = 36 + {p^2} - 12p$
Subtracting ${p^2}$ from both sides, we get
$\begin{array}{l} \Rightarrow {p^2} + {q^2} - {p^2} = 36 + {p^2} - 12p - {p^2}\\\ \Rightarrow {q^2} = 36 - 12p\end{array}$
Factoring out $- 12$ from the expression, we get
$\Rightarrow {q^2} = - 12\left( {p - 3} \right)$
We will replace $p$ with $x$, and $q$ with $y$ to find the locus of the point $P\left( {p,q} \right)$.
Therefore, we get the locus of the point as
$\Rightarrow {y^2} = - 12\left( {x - 3} \right)$
Therefore, the locus of the given point is ${y^2} = - 12\left( {x - 3} \right)$.
Thus, the correct option is option (a).

Note: We used the algebraic identity to simplify the expression ${\left( {6 - p} \right)^2}$ to $36 + {p^2} - 12p$. The square of the difference of two numbers is given by the algebraic identity ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$. Here we had to find the locus of points. A locus is defined as the set of all the points which form a surface or a curve and satisfy a particular condition.