A points 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
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} (1 - e^{2})} = 1$

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Solution

Suppose that the co-ordinates of the point are (x , y), then the given condition
$\sqrt{[{(x - a e)}^{2} + y^{2}]} + \sqrt{[{(x + a e)}^{2} + y^{2}]} = 2 a$ .........(1)
Now
$[{(x - a e)}^{2} + y^{2}] - [{(x + a e)}^{2} + y^{2}] = - 4 a e x$ ............(2)
$[∵ (a - b)^{2} - (a + b)^{2} = - 4 a b]$
On dividing (2) by (1), we get
$\sqrt{[{(x - a e)}^{2} + y^{2}]} - \sqrt{[{(x + a e)}^{2} + y^{2}]} = - 2 e x$
$[∵ \frac{L - M}{(\sqrt{L}) + (\sqrt{M})} = (\sqrt{L}) - (\sqrt{M})]$
Adding (1) and (3) we get
$2 \sqrt{[{(x - a e)}^{2} + y^{2}]} = 2 (a - e x)$ . Square
$x^{2} - 2 a e x + a^{2} e^{2} + y^{2} = a^{2} - 2 a e x + e^{2} x^{2}$
or $x^{2} (1 - e^{2}) + y^{2} = a^{2} (1 - e^{2})$
or $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} (1 - e^{2})} = 1$
Note : The above method will be referred to as L-M method throughout this book.

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A points 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 x2a2+y2a2(1−e2)=1

A points 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
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} (1 - e^{2})} = 1$