[ x2/a2 ] + [ y2/b2 ] = 1
Let x'ox & yoy' be the co-ordinate axes.
Let F(c, o) and f'(- c, o) be 2 given fixed points .
Let us consider the locus of a point which moves in such a way that the sum of its distances from F & F' remains constant say equal to 2a where a > c.
Let P(x, y) be any point on the locus.
Then
PF + PF' = 2a
=> √[(x - c)2 + y2] + √[(x + c)2 + y2] = 2a
√[(x + c)2 + y2] = 2a - √[(x-c)2 + y2]
On squaring both sides,
we get
[(x + c)2 + y2] = 4a2 + (x - c)2 + y2 - 4a√[(x - c)2 + y2]
[(x + c) 2 - (x - c) 2] - 4a2 = - 4a √[(x - c)2 + y2]
4 x c – 4a2 = - 4a √[(x - c)2 + y2]
√[(x - c)2 + y2] = a – (c/a) x
Again squaring on both sides ,
we get
(x - c)2 + y2 = a2 + [c2x2/a2] - 2cx
x2 - [c2x2/a2] + y2 = a2 - c2
x2 [1 – (c2/a2)] + y2 = a2 - c2
[x2(a2 - c2)] / a2 +y2 = a2 - c2
Dividing by (a2 - c2)
we get
(x2/a2) + (y2/(a2- c2)) = 1
(x2/a2) + (y2/b2) = 1 , where b2 = a2 - c2
Thus (x2/a2) + (y2/b2) = 1 is the required equation of an ellipse in standard form