In the following figure, the line that touches the ellipse at only one point, called A, is the tangent line to the ellipse at that point. C is the center of the ellipse. Point $L'$ is the point where the perpendicular passing through C to the tangent line intersects the tangent line. Point L, instead, is the intersection between this perpendicular and the ellipse.
I know that the ratio $\frac{AL'}{L'C}$ is given. I would like to calculate the length of $AC$. I thought I could calculate $L'C$ (and then $AC$) using question Distance point on ellipse to centre; however, by doing so, I can only calculate $LC$. Any suggestions?
Best Answer
We may suppose that the equation of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $A(a\cos t,b\sin t)$ and $C(0,0)$.
Then, the equation of the line $AL'$ is given by $ \frac{x\cos t}{a}+\frac{y\sin t}{b}=1$. So, we can write $L'(c,\frac{ab-bc\cos t}{a\sin t})$. Since $CL'$ is perpendicular to $AL'$, solving $$\frac{ab-bc\cos t}{ac\sin t}\times \frac{-b\cos t}{a\sin t}=-1$$ for $c$ gives $c=\frac{ab^2\cos t}{a^2\sin^2t+b^2\cos^2t}$. So, we have $L'(\frac{ab^2\cos t}{a^2\sin^2t+b^2\cos^2t},\frac{a^2b\sin t}{a^2\sin^2t+b^2\cos^2t})$.
Since $L$ is on the line segment $CL'$, we can write $L(L'_xd,L'_yd)$ where $0\lt d\lt 1$.
Since $L$ is on the ellipse, solving $\frac{(L'_xd)^2}{a^2}+\frac{(L'_yd)^2}{b^2}=1$ for $d$ gives $d=\frac{a^2\sin^2t+b^2\cos^2t}{\sqrt{a^4\sin^2t+b^4\cos^2t}}$.
So, we have $L(\frac{ab^2\cos t}{\sqrt{a^4\sin^2t+b^4\cos^2t}},\frac{a^2b\sin t}{\sqrt{a^4\sin^2t+b^4\cos^2t}})$, and so $$AL'=\frac{|(a^2-b^2)\cos t\sin t|}{\sqrt{a^2\sin^2t+b^2\cos^2t}}$$ $$L'C=\frac{ab}{\sqrt{a^2\sin^2t+b^2\cos^2t}}$$
Now, if we know $r:=\frac{AL'}{L'C}$, then we have $r=\frac{|a^2-b^2|}{ab}|\cos t\sin t|=\frac{|a^2-b^2|}{2ab}|\sin(2t)|$ which implies $$|\sin(2t)|=\frac{2abr}{|a^2-b^2|}$$ $$1-\cos^2(2t)=\frac{4a^2b^2r^2}{(a^2-b^2)^2}$$ $$\cos^2(2t)=1-\frac{4a^2b^2r^2}{(a^2-b^2)^2}$$ $$\cos(2t)=\pm\sqrt{1-\frac{4a^2b^2r^2}{(a^2-b^2)^2}}$$
We can represent $AC$ by $\cos(2t)$ as follows : $$\begin{align}AC&=\sqrt{a^2\cos^2t+b^2\sin^2t} \\\\&=\sqrt{a^2\cos^2 t+b^2(1-\cos^2t)} \\\\&=\sqrt{b^2+(a^2-b^2)\cos^2t} \\\\&=\sqrt{b^2+(a^2-b^2)\frac{1+\cos(2t)}{2}} \\\\&=\sqrt{\frac{a^2+b^2}{2}+\frac{a^2-b^2}2\cos(2t)}\end{align}$$
Therefore, we finally get $$\color{red}{AC=\sqrt{\frac{a^2+b^2}{2}\pm \frac{a^2-b^2}2\sqrt{1-\frac{4a^2b^2}{(a^2-b^2)^2}\bigg(\frac{AL'}{L'C}\bigg)^2}}}$$