Question

A point moves that the sum of its distances from two fixed points \[( - ae,0)\] and \[(ae,0)\] is always \[2a\]. Prove that the equation of the locus is \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{a^2}(1 - {e^2})}} = 1\] .

Answer 1

Hint: We will assume the moving point to be \[\left( {x,y} \right)\] and then by using the distance formula ,we will try to calculate the distance of the moving point from \[( - ae,0)\]and \[(ae,0)\] then we will use the given condition that the sum of these distances is always \[2a\] to get an equation correlating \[x\] and \[y\].By solving this equation ,we can get the locus of moving point.

Formula used
If there are two points A and B having coordinates \[({x_1},{y_1})\] and \[({x_2},{y_2})\] then distance between these points (AB) is given by
   AB =\[\sqrt {{{({x_1} - {x_2})}^2} + {{({y_2} - {y_1})}^2}} \]

Complete step by step solution:
Given points are \[( - ae,0)\]and \[(ae,0)\].Let the coordinate of moving point be \[\left( {x,y} \right)\].Then we can calculate the distance between the moving point and the given points.
Using the distance formula \[\sqrt {{{({x_1} - {x_2})}^2} + {{({y_2} - {y_1})}^2}} \]to find the distance between point \[\left( {x,y} \right)\]and \[( - ae,0)\]
Distance \[{d_1} = \sqrt {{{(x - ( - ae))}^2} + {{(y - 0)}^2}} \]
 \[ \Rightarrow {d_1} = \sqrt {{{(x + ae)}^2} + {y^2}} \]
Again, using the distance formula to find the distance between \[\left( {x,y} \right)\]and \[(ae,0)\]
 Distance \[{d_2} = \sqrt {{{(x - ae)}^2} + {{(y - 0)}^2}} \]
 \[ \Rightarrow {d_2} = \sqrt {{{(x - ae)}^2} + {y^2}} \]
  Given- sum of the distances \[{d_1}\] and \[{d_2}\] is \[2a\].
     \[ \Rightarrow {d_1} + {d_2} = 2a\]
Substituting the values \[{d_1} = \sqrt {{{(x + ae)}^2} + {y^2}} \]and \[{d_2} = \sqrt {{{(x - ae)}^2} + {y^2}} \]
\[\sqrt {{{(x + ae)}^2} + {y^2}} + \sqrt {{{(x - ae)}^2} + {y^2}} = 2a\]
Rearranging the terms
\[\sqrt {{{(x + ae)}^2} + {y^2}} = 2a - \sqrt {{{(x - ae)}^2} + {y^2}} \]
Squaring on both sides
\[\begin{array}{l}{\left( {\sqrt {{{(x + ae)}^2} + {y^2}} } \right)^2} = {\left( {2a - \sqrt {{{(x - ae)}^2} + {y^2}} } \right)^2}\\ \Rightarrow \left( {{x^2} + {a^2}{e^2} + 2aex + {y^2}} \right) = 4{a^2} + \left( {{x^2} - 2aex + {a^2}{e^2} + {y^2}} \right) - 4a\sqrt {{{(x - ae)}^2} + {y^2}} \\ \Rightarrow 4a\sqrt {{{(x - ae)}^2} + {y^2}} + 4aex = 4{a^2}\end{array}\]
Dividing by \[4a\] on both sides
\[\sqrt {{{(x - ae)}^2} + {y^2}} + ex = a\]\[\]
Subtracting \[ex\] from both sides
\[\sqrt {{{(x - ae)}^2} + {y^2}} = a - ex\]
Again, squaring both sides of the equation
\[{\left( {\sqrt {{{(x - ae)}^2} + {y^2}} } \right)^2} = {\left( {a - ex} \right)^2}\]
Simplifying the equation
\[{x^2} + {a^2}{e^2} - 2aex + {y^2} = {a^2} + {e^2}{x^2} - 2aex\]
\[ \Rightarrow {x^2} + {a^2}{e^2} + {y^2} = {a^2} + {e^2}{x^2}\]
\[ \Rightarrow {x^2} - {e^2}{x^2} + {y^2} = {a^2} - {a^2}{e^2}\]
\[ \Rightarrow {x^2}\left( {1 - {e^2}} \right) + {y^2} = {a^2}(1 - {e^2})\]
Dividing by \[{a^2}(1 - {e^2})\] on both sides
\[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{a^2}(1 - {e^2})}} = 1\]

Hence proved that the locus of the point is \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{a^2}(1 - {e^2})}} = 1\].

Note: In this question, students may get stuck at some step if they try to square the equations with under roots on both sides. In this case, the degree of variables will go up to 4 and it will become too long an expansion to solve. Hence, the easier way is to shift the one radical term to the other side and then square the equation to get rid of the first radical term. The second radical term can be removed by squaring the equation again and this will ease the calculation.