Question

Three distinct points are given in the 2-dimensional coordinate plane such that the ratio of the distance of any one of them from the point (1, 0) to the distance from the point (-1, 0) is equal to $\dfrac{1}{3}$. Then the locus of point A is,
$\left( a \right){x^2} - {y^2} - \dfrac{5}{2}x + 1 = 0$
$\left( b \right){x^2} + {y^2} - \dfrac{5}{2}x + 1 = 0$
$\left( c \right){x^2} + {y^2} + \dfrac{5}{2}x + 1 = 0$
$\left( d \right)$ None of these.

Answer 1

Hint: In this particular question use the concept that the distance between two points $\left( {{x_1},{y_1}} \right){\text{ and }}\left( {{x_2},{y_2}} \right)$ is given as $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $ so use this concept to reach the solution of the question.

Complete step-by-step solution:
Let us consider the three distinct points in 2- dimensional be P, Q, and R.
Let L = (1, 0) and M = (-1, 0)
Now according to the question, the ratio of the distance of any one of them from the point (1, 0) to the distance from the point (-1, 0) is equal to $\dfrac{1}{3}$.
$ \Rightarrow \dfrac{{{D_{PL}}}}{{{D_{PM}}}} = \dfrac{1}{3},\dfrac{{{D_{QL}}}}{{{D_{QM}}}} = \dfrac{1}{3},\dfrac{{{D_{RL}}}}{{{D_{RM}}}} = \dfrac{1}{3}$
Now let a point A (h, k) moves in such a way that,
$ \Rightarrow \dfrac{{{D_{AL}}}}{{{D_{AM}}}} = \dfrac{1}{3}$................ (1)
Therefore, P, Q, and R will lie on a locus of point A.
Now as we know that the distance between two points $\left( {{x_1},{y_1}} \right){\text{ and }}\left( {{x_2},{y_2}} \right)$ is given as $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
Now let, (h, k) = $\left( {{x_1},{y_1}} \right)$, L = (1, 0) = $\left( {{x_2},{y_2}} \right)$ and M = (-1, 0) = $\left( {{x_3},{y_3}} \right)$
So, ${D_{AL}} = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} = \sqrt {{{\left( {1 - h} \right)}^2} + {{\left( {0 - k} \right)}^2}} $
And ${D_{AM}} = \sqrt {{{\left( {{x_3} - {x_1}} \right)}^2} + {{\left( {{y_3} - {y_1}} \right)}^2}} = \sqrt {{{\left( { - 1 - h} \right)}^2} + {{\left( {0 - k} \right)}^2}} $
Now substitute these values in equation (1) we have,
$ \Rightarrow \dfrac{{\sqrt {{{\left( {1 - h} \right)}^2} + {{\left( {0 - k} \right)}^2}} }}{{\sqrt {{{\left( { - 1 - h} \right)}^2} + {{\left( {0 - k} \right)}^2}} }} = \dfrac{1}{3}$
\[ \Rightarrow \dfrac{{\sqrt {{{\left( {1 - h} \right)}^2} + {k^2}} }}{{\sqrt {{{\left( {1 + h} \right)}^2} + {k^2}} }} = \dfrac{1}{3}\]
Now squaring on both sides we have,
\[ \Rightarrow \dfrac{{{{\left( {1 - h} \right)}^2} + {k^2}}}{{{{\left( {1 + h} \right)}^2} + {k^2}}} = {\left( {\dfrac{1}{3}} \right)^2} = \dfrac{1}{9}\]
\[ \Rightarrow 9{\left( {1 - h} \right)^2} + 9{k^2} = {\left( {1 + h} \right)^2} + {k^2}\]
Now expand the square according to the property ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ so we have,
\[ \Rightarrow 9\left( {1 + {h^2} - 2h} \right) + 9{k^2} = \left( {1 + {h^2} + 2h} \right) + {k^2}\]
Now simplify it we have,
\[ \Rightarrow 9 + 9{h^2} - 18h + 9{k^2} = 1 + {h^2} + 2h + {k^2}\]
\[ \Rightarrow 8 + 8{h^2} - 20h + 8{k^2} = 0\]
Now divide by 8 throughout we have,
\[ \Rightarrow 1 + {h^2} - \dfrac{{20}}{8}h + {k^2} = 0\]
\[ \Rightarrow {h^2} + {k^2} - \dfrac{5}{2}h + 1 = 0\]
Now in place of (h, k) substitute (x, y) we have,
\[ \Rightarrow {x^2} + {y^2} - \dfrac{5}{2}x + 1 = 0\]
So this is the required locus of point A.
Hence option (b) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the distance formula between two points which is stated above, so first find out the distance of the point (1, 0) from the point (h, k) and then find the distance of the point (-1, 0) from the point (h, k) and the ratio of these two distance is given as 1:3 so equate them and simplify we will get the required locus of the point A.