Solution:
Let P (h, k) represent any point on the locus
And let the coordinates of A and B be given by (ae, 0) and(-ae, 0).
Where, we have:
PA – PB = 2a
Upon squaring both sides we get:
$ {{\left( \mathbf{eh}\text{ }+\text{ }\mathbf{a} \right)}^{\mathbf{2}}}~=\text{ }{{\left( \mathbf{h}\text{ }+\text{ }\mathbf{ae} \right)}^{\mathbf{2}}}~+\text{ }{{\left( \mathbf{k}-\mathbf{0} \right)}^{\mathbf{2}}} $
$ {{\mathbf{e}}^{\mathbf{2}}}{{\mathbf{h}}^{\mathbf{2}}}~+\text{ }{{\mathbf{a}}^{\mathbf{2}}}~+\text{ }\mathbf{2aeh}\text{ }=\text{ }{{\mathbf{h}}^{\mathbf{2}}}~+\text{ }{{\mathbf{a}}^{\mathbf{2}}}{{\mathbf{e}}^{\mathbf{2}}}~+\text{ }\mathbf{2aeh}\text{ }+\text{ }{{\mathbf{k}}^{\mathbf{2}}}{{\mathbf{h}}^{\mathbf{2}}}~ $
$ \left( {{\mathbf{e}}^{\mathbf{2}}}-\mathbf{1} \right)-{{\mathbf{k}}^{\mathbf{2}}}~=\text{ }{{\mathbf{a}}^{\mathbf{2}}}~({{\mathbf{e}}^{\mathbf{2}}}-\mathbf{1}) $
Upon replacing (h, k) with (x, y)
The locus of a point such that the difference of its distances from (ae, 0) and (-ae, 0) is 2a.
Where b2 = a2 (e2 – 1)
Hence proved.