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Figure Diagram

I would like to know how to find $a$ and $b$ of an ellipse if I know $K$, $L$ and $m$ as shown in the figure .

I can find the answer to this using the equations below and an Excel Goal Seek or Python but I'd be interested to see how to solve this directly. I get stuck trying to simplify the simultaneous equations.

$$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ \\ \frac{L^2}{a^2}+\frac{(-b+K)^2}{b^2}=1\quad(1)\\ \\ \\ \\ y= mx\pm \sqrt{a^2m^2+b^2}\\ \\ -b+K=mL\pm\sqrt{a^2m^2+b^2}\;\;\;\;(2)\\ $$

Rearranging $(1)$ $$ a=\sqrt{\frac{L^2}{1-\frac{(-b+K)^2}{b^2}}} $$

Thanks in Advance!

Rohit Singh
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Sam Williams
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    There seems to be some inaccuracies with the figure. If $b$ and $K$ are both positive constants, then the y coordinate of the intersection should be $K-b$. I believe that’s what you meant as that’s also what you used in your calculations. – person Apr 29 '22 at 15:34
  • Yes @person my figure is wrong, thanks for pointing it out. – Sam Williams May 03 '22 at 14:49

1 Answers1

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Let a tangency point be $P=(x_0,y_0)$ and suppose the normal at $P$ intersects $x$-axis at $K=(d,0)$ (if $m$ is the slope of the tangent, then $d=my_0+x_0$). Foci $F$ and $F'$ lie on $x$-axis at a distance $c$ from the origin and normal $PK$ is the bisector of $\angle FPF'$. Hence: $$ FK:F'K=FP:F'P, $$ that is: $$ (c-d):(c+d)=\sqrt{(c-x_0)^2+y_0^2}:\sqrt{(c+x_0)^2+y_0^2}. $$ Squaring both sides and simplifying, we can find $c$: $$ c^2={d(x_0^2+y_0^2-dx_0)\over x_0-d}. $$ Finally, once we know $c$ we can compute semi-axes: $2a=PF+PF'$ and $b=\sqrt{a^2-c^2}$.

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Intelligenti pauca
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