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

Hint: 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.