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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.

neslrac
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1 Answers1

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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 $

  • Amazing, thanks a lot. Your $h$ and my $x_0$ have the same assignments btw.: I could have eliminated $2C$ after leaving out the zero terms. I feel this is the right way to do it, but maybe not for me - I'll use the secret chainsaw. – neslrac Nov 29 '22 at 23:49