TITLE: Find standard form parameters for ellipse
QUESTION [0 upvotes]: I got this ellipse yesterday with center on x-axis going through points (0, p) and (0, -p) and touching unit circle twice (double contact) at x = t
$$
x^2+y^2-1+\dfrac{1-p^2}{t^2}(x-t)^2=0
$$
Here is the context when I asked for the solution Perspective view of longitudinal great circles - ellipses inside a circle
I changed the variable names to $p$ for pole and $t$ for touching point to clarify that this ellipse equation does not contain the semis $a$ and $b$ or the center directly.
Now how to determine center and semi-axis? What I finally did after tons of mistakes is multiply out etc. and put it into general equation, keeping the variables/parameters:
$$
(1 + \dfrac{1-p^2}{t^2}) x^2+y^2 - \dfrac{2(1-p^2)}{t}x - p^2 = 0
$$
I have no $xy$ ("B") or $y$ ("E") in this case, so the long converting equations on wikipedia/ellipse shrink a bit, e.g.
$$
x_0 = \dfrac{2CD}{-4AC}
$$
I this overkill? The semi-axis have a much longer equation.
Because the person who offered the solution guided me to "completing the square". Maybe he was thinking of a specific value pair for $p$ and $t$. But even so this method seems cumbersome, because I don't have nice coefficients, and also I want to find several ellipses with different $p$ and $t$. I must also add/admit that I never heard of this method.
To me it seems these general-form-coefficients to canonical-form-parameters equations on wikipedia are made exactly for a case like this. Now I can go: with given $p$ and $t$ I get such and such A, C, D and F, and thus my $x_0$, $a$ and $b$ are this.

REPLY [3 votes]: Your ellipse equation is
$$
(1 + \dfrac{1-p^2}{t^2}) x^2+y^2 - \dfrac{2(1-p^2)}{t}x - p^2 = 0
$$
Let $ A = (1 + \dfrac{1-p^2}{t^2}) $ and $ B = - \dfrac{2(1-p^2)}{t} $
Then the equation becomes
$ A x^2 + y^2 + B x = p^2 $
Completing the square in $x$ gives
$ A (x + \dfrac{B}{2A} )^2 - \dfrac{B^2}{4 A} + y^2 = p^2 $
Let $ h = - \dfrac{B}{2A} $ , and $ C = p^2 + \dfrac{B^2}{4A} $
Then the equation now becomes
$ A (x - h)^2 + y^2 = C $
Divide through by $C$
$ \bigg( \dfrac{A}{C} \bigg) (x - h)^2 + \dfrac{1}{C} y^2 = 1 $
So the semi-axes are
$ a =  \sqrt{ \dfrac{C}{A} } $ , and $ b = \sqrt{C} $
And the now the equation is
$ \dfrac{(x - h)^2}{a^2} + \dfrac{y^2}{b^2} = 1 $
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