Calculating the Distance from a Point on the Tangent to an Ellipse to the Center

Question

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?

Answer 1

Let's WLOG take the centre of the ellipse be origin. Then the ellipse can be described as $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$ We state without a proof a lemma that

The equation of tangent to the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ at $(x_1,y_1)$ is $\dfrac{x_1x}{a^2}+\dfrac{y_1y}{b^2}=1$.

This is a well-known result from conic section, you can ask for a proof if you need. Now given this fact, if we know the coordinates of $L'=(l_1,l_2)$, then we have $$b^2(x_1l_1)+a^2(y_1l_2)=a^2b^2\iff x_1=\dfrac{a^2(b^2-y_1l_2)}{b^2l_1}$$ However, we know that $(x_1,y_1)$ is point on the ellipse, so we get a quadratic equation $$\dfrac{a^2(b^2-y_1l_2)^2}{b^4l_1^2}+\dfrac{y_1^2}{b^2}=1$$ So you can solve for $A=(x_1,y_1)$. And you get $AL$, and thus $LC$ by the ratio.

Now if you are given $A$, then you get the slope of $CA$, and the given information can be described as $\tan\angle ACL$, so you can find $\angle ACL$. Combined with the slope of $CA$, you can get the slope of $LC$, thus you can solve the intersecting of the line of tangent at $A$ and the line $LC$, you get the coordinates of $L$, thus $LC$.

Answer 2

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 $$\color{red}{AC=\sqrt{\frac{a^2+b^2}{2}\pm \frac{a^2-b^2}2\sqrt{1-\frac{4a^2b^2}{(a^2-b^2)^2}\bigg(\frac{AL'}{L'C}\bigg)^2}}}$$

Answer 3

The problem data is incomplete, because for any $\lambda>0$ if we do the homothety with the center $C$ and the coefficient $\lambda$ of the whole construction then the ratio $k=\frac{AL'}{L'C}$ is not changed, whereas $|AC|$ is multiplied by $\lambda$.

To calculate the respective values, let us follow Hosam Hajeer's suggestion, introduce Cartesian coordinates on the plane such that the ellipse has a canonical parametric equation $(a \cos\varphi, b\sin\varphi)$ for some $a,b>0$ and $\varphi\in\ [0,2\pi]$. Then the point $C$ is the origin $(0,0)$. Let the point $A$ has the coordinates $(a\cos\alpha, b\sin\alpha)$. So $|AC|^2=a^2\cos^2\alpha+b^2\sin^2\alpha$. The line $L'C$ is collinear to the vector $(b\cos\alpha, a\sin\alpha)$. Since $AL'$ is perpendicular to $L'C$, $k=\tan\angle ACL$ (that is, we consider $k$ to be the signed ratio). For the simplicity suppose that $\alpha\not\in\frac\pi{2}\mathbb Z$. Pick the point $X(0,1)$ from the $Ox$ axis. Then $\tan\angle ACX=\frac{a}{b\tan\alpha}$, $\tan\angle LCX=\frac{b}{a\tan\alpha}$, and $$k=\tan\angle ACL=\tan(\angle LCX-\angle ACX)=$$ $$\frac{\tan \angle LCX-\tan \angle ACX}{1+\tan \angle ACX\cdot \tan\angle LCX}=\frac{\frac{b}{a}-\frac{a}{b}}{\tan\alpha+\tan^{-1}\alpha}.$$