Question

What is the locus of a point whose difference of distance from points \[\left( { \pm 3,0} \right)\] is 4?
A. \[\dfrac{{{x^2}}}{4} - \dfrac{{{y^2}}}{5} = 1\]
B. \[\dfrac{{{x^2}}}{5} - \dfrac{{{y^2}}}{4} = 1\]
C. \[\dfrac{{{x^2}}}{2} - \dfrac{{{y^2}}}{3} = 1\]
D. \[\dfrac{{{x^2}}}{3} - \dfrac{{{y^2}}}{2} = 1\]

Answer 1

Hint: First we will assume the coordinate the point whose locus we need to find. Then we will apply the distance formula of two points to find the distance of the point from the points \[\left( { \pm 3,0} \right)\]. Then equate the difference of distances to 4.

Formula used:
The distance between the points \[d = \sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}} \]
Algebraical identity: \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]

Complete step by step solution:
Assume the coordinate of the point is \[\left( {x,y} \right)\].
Apply the distance formula to find the distance between \[\left( {x,y} \right)\] and \[\left( {3,0} \right)\].
The distance between \[\left( {x,y} \right)\] and \[\left( {3,0} \right)\] is \[{d_1} = \sqrt {{{\left( {x - 3} \right)}^2} + {{\left( {y - 0} \right)}^2}} \]
\[ \Rightarrow {d_1} = \sqrt {{{\left( {x - 3} \right)}^2} + {y^2}} \]
Apply the distance formula to find the distance between \[\left( {x,y} \right)\] and \[\left( { - 3,0} \right)\].
The distance between \[\left( {x,y} \right)\] and \[\left( { - 3,0} \right)\] is \[{d_2} = \sqrt {{{\left( {x - \left( { - 3} \right)} \right)}^2} + {{\left( {y - 0} \right)}^2}} \]
\[ \Rightarrow {d_2} = \sqrt {{{\left( {x + 3} \right)}^2} + {y^2}} \]
Now find the difference between the distances:
\[{d_2} - {d_1} = \sqrt {{{\left( {x + 3} \right)}^2} + {y^2}} - \sqrt {{{\left( {x - 3} \right)}^2} + {y^2}} \]
Equate the difference of distance to 4:
\[\sqrt {{{\left( {x + 3} \right)}^2} + {y^2}} - \sqrt {{{\left( {x - 3} \right)}^2} + {y^2}} = 4\]
Take square both sides
\[ \Rightarrow {\left( {\sqrt {{{\left( {x + 3} \right)}^2} + {y^2}} - \sqrt {{{\left( {x - 3} \right)}^2} + {y^2}} } \right)^2} = {4^2}\]
Apply the algebraical identity:
\[ \Rightarrow {\left( {\sqrt {{{\left( {x + 3} \right)}^2} + {y^2}} } \right)^2} + {\left( {\sqrt {{{\left( {x - 3} \right)}^2} + {y^2}} } \right)^2} - 2\sqrt {{{\left( {x + 3} \right)}^2} + {y^2}} \sqrt {{{\left( {x - 3} \right)}^2} + {y^2}} = 16\]
\[ \Rightarrow {\left( {x + 3} \right)^2} + {y^2} + {\left( {x - 3} \right)^2} + {y^2} - 2\sqrt {{{\left( {x + 3} \right)}^2} + {y^2}} \sqrt {{{\left( {x - 3} \right)}^2} + {y^2}} = 16\]
\[ \Rightarrow {x^2} + 6x + 9 + {y^2} + {x^2} - 6x + 9 + {y^2} - 2\sqrt {{{\left( {x + 3} \right)}^2} + {y^2}} \sqrt {{{\left( {x - 3} \right)}^2} + {y^2}} = 16\]
\[ \Rightarrow 2{x^2} + 2{y^2} + 18 - 2\sqrt {{{\left( {x + 3} \right)}^2} + {y^2}} \sqrt {{{\left( {x - 3} \right)}^2} + {y^2}} = 16\]
Rewrite the equation:
\[ \Rightarrow 2{x^2} + 2{y^2} + 18 - 16 = 2\sqrt {{{\left( {x + 3} \right)}^2} + {y^2}} \sqrt {{{\left( {x - 3} \right)}^2} + {y^2}} \]
\[ \Rightarrow 2{x^2} + 2{y^2} + 2 = 2\sqrt {{{\left( {x + 3} \right)}^2} + {y^2}} \sqrt {{{\left( {x - 3} \right)}^2} + {y^2}} \]
Take common 2 from the left side expression:
\[ \Rightarrow 2\left( {{x^2} + {y^2} + 1} \right) = 2\sqrt {{{\left( {x + 3} \right)}^2} + {y^2}} \sqrt {{{\left( {x - 3} \right)}^2} + {y^2}} \]
Cancel out 2 from left side and right-side expression
\[ \Rightarrow \left( {{x^2} + {y^2} + 1} \right) = \sqrt {{{\left( {x + 3} \right)}^2} + {y^2}} \sqrt {{{\left( {x - 3} \right)}^2} + {y^2}} \]
Take square both sides
\[ \Rightarrow {\left( {{x^2} + {y^2} + 1} \right)^2} = {\left( {\sqrt {{{\left( {x + 3} \right)}^2} + {y^2}} \sqrt {{{\left( {x - 3} \right)}^2} + {y^2}} } \right)^2}\]
Apply the algebraical identity:
\[ \Rightarrow {x^4} + {y^4} + 1 + 2{x^2} + 2{y^2} + 2{x^2}{y^2} = \left[ {{{\left( {x + 3} \right)}^2} + {y^2}} \right]\left[ {{{\left( {x - 3} \right)}^2} + y} \right]\]
\[ \Rightarrow {x^4} + {y^4} + 1 + 2{x^2} + 2{y^2} + 2{x^2}{y^2} = \left[ {{x^2} + 6x + 9 + {y^2}} \right]\left[ {{x^2} - 6x + 9 + {y^2}} \right]\]
\[ \Rightarrow {x^4} + {y^4} + 1 + 2{x^2} + 2{y^2} + 2{x^2}{y^2} = {\left( {{x^2} + 9 + {y^2}} \right)^2} - {\left( {6x} \right)^2}\]
\[ \Rightarrow {x^4} + {y^4} + 1 + 2{x^2} + 2{y^2} + 2{x^2}{y^2} = {x^4} + 81 + {y^4} + 18{x^2} + 18{y^2} + 2{x^2}{y^2} - 36{x^2}\]
\[ \Rightarrow {x^4} + {y^4} + 1 + 2{x^2} + 2{y^2} + 2{x^2}{y^2} - {x^4} - 81 - {y^4} - 18{x^2} - 18{y^2} - 2{x^2}{y^2} + 36{x^2} = 0\]
\[ \Rightarrow 20{x^2} - 16{y^2} - 80 = 0\]
\[ \Rightarrow 20{x^2} - 16{y^2} = 80\]
Divide both sides by 80
\[ \Rightarrow \dfrac{{20{x^2}}}{{80}} - \dfrac{{16{y^2}}}{{80}} = \dfrac{{80}}{{80}}\]
\[ \Rightarrow \dfrac{{{x^2}}}{4} - \dfrac{{{y^2}}}{5} = 1\]
The locus of the point is \[\dfrac{{{x^2}}}{4} - \dfrac{{{y^2}}}{5} = 1\].
Hence option A is the correct option.

Note: Students often don’t understand the locus. The locus of a point is generally a geometrical shape. In the given question, we get a hyperbola.