How to get from $x^2 + y^2 = e^2(d + x)^2$ to $(\frac{(1-e^2)^2}{e^2d^2})(x-\frac{e^2d}{1-e^2})^2+(\frac{(1-e^2)}{e^2d^2})y^2=1$ I'm working through a problem in my pre-calc book and I don't know how to get the equation into the same form as the book.  We start with $x^2 + y^2 = e^2(d + x)^2$ and the book gets to the following form:
$(\frac{(1-e^2)^2}{e^2d^2})(x-\frac{e^2d}{1-e^2})^2+(\frac{(1-e^2)}{e^2d^2})y^2=1$
When I try I get stuck.  Here are my steps:
1.) $x^2 + y^2 = e^2(d + x)^2$
2.) $x^2+y^2 = e^2(d^2 + dx + x^2)$
3.) $x^2+y^2 = e^2d^2 + e^2dx + e^2x^2$
4.) $x^2 - e^2x^2 - e^2dx = e^2d^2 -y^2$ (Grouping like terms)
5.) $(1-e^2)x^2 - e^2dx= e^2d^2 - y^2$
6.) $x^2 - \frac{e^2dx}{(1-e^2)}= \frac{e^2d^2 - y^2}{(1-e^2)}$
7.) $x^2 - \frac{e^2dx}{1-e^2} + (\frac{e^2d}{2(1-e^2)})^2 = \frac{e^2d^2 - y^2}{(1-e^2)} + (\frac{e^2d}{2(1-e^2)})^2$ Completing the square.
8.) $(x^2 - \frac{e^2d}{2(1-e^2)})^2 = \frac{e^2d^2 - y^2}{(1-e^2)} + (\frac{e^2d}{2(1-e^2)})^2$
But from here I don't know where to go.  Any help is Much appreciated.

To be completed

*

*given equation

*you missed a term $dx$ inside the bracket on the RHS

*$x^2+y^2 = e^2d^2 + 2e^2dx + e^2x^2$

*$x^2 - e^2x^2 - 2e^2dx = e^2d^2 -y^2$  (same logic with coefficient corrected)

*$(1-e^2)x^2 - 2e^2dx= e^2d^2 - y^2$

*$$x^2 - \frac{2e^2dx}{(1-e^2)}= \frac{e^2d^2 - y^2}{(1-e^2)}$$

*$$x^2 - \frac{2e^2dx}{1-e^2} + (\frac{e^2d}{(1-e^2)})^2 = \frac{e^2d^2 - y^2}{(1-e^2)} + (\frac{e^2d}{(1-e^2)})^2$$ (same logic here)

*The first term inside the bracket on the LHS should be $x$.  $$(x - \frac{e^2d}{(1-e^2)})^2 = \frac{e^2d^2 - y^2}{(1-e^2)} + (\frac{e^2d}{(1-e^2)})^2$$
Can you continue?
