Problem source

The straight line $OSA,$ where $O$ is the origin, bisects the angle between the positive $x$ and $y$ axes. The ellipse $E$ has $S$ as focus. In polar coordinates with $S$ as pole and $SA$ as the initial line, $E$ has equation $\ell=r(1+e\cos\theta).$ Show that, at the point on $E$ given by $\theta=\alpha,$ the gradient of the tangent to the ellipse is given by 
\[
\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\sin\alpha-\cos\alpha-e}{\sin\alpha+\cos\alpha+e}.
\]
The points on $E$ given by $\theta=\alpha$ and $\theta=\beta$ are the ends of a diameter of $E$. Show that 
\[
\tan(\alpha/2)\tan(\beta/2)=-\frac{1+e}{1-e}.
\]
[\textbf{Hint. }A diameter of an ellipse is a chord through its centre.]

Solution source

\begin{center}
    \begin{tikzpicture}
        \def\xl{-4};
        \def\xu{4};
    
        \draw[->] (-4,0) -- (4,0);
        \draw[->] (0,-4) -- (0,4);


        \draw[->] (1,1) -- (4,4);

        

        \filldraw (2,3) circle (1pt) node[above right] {$(x,y)$};
        \filldraw (1,1) circle (1.5pt) node[below] {$S$};

    \end{tikzpicture}
\end{center}

\begin{align*}
&& \ell &= r(1 + e \cos \theta) \\
\Rightarrow && 0 &= \frac{\d r}{\d \theta}(1 + e \cos \theta) - re \sin \theta \\
\Rightarrow && \frac{\d r}{\d \theta} &= \frac{re \sin \theta}{1+e \cos \theta}
\end{align*}


Suppose we consider the $(x',y')$ plane, which is essentially the $x-y$ plan rotated by $45^\circ$, then we would have

\begin{align*}
&& \frac{\d y'}{ \d x'} &= \frac{\frac{\d y'}{\d \theta}}{\frac{\d x'}{\d \theta}} \\
&&&= \frac{\frac{\d r}{\d \theta} \sin \theta + r\cos \theta}{\frac{\d r}{\d \theta} \cos \theta - r\sin \theta} \\
&&&= \frac{\frac{re \sin \theta}{1+e \cos \theta} \sin \theta + r\cos \theta}{\frac{re \sin \theta}{1+e \cos \theta} \cos\theta -r\sin \theta} \\
&&&= \frac{re\sin^2 \theta+r \cos \theta(1+e \cos \theta)}{re\sin \theta \cos \theta -r \sin \theta (1+e \cos \theta)} \\
&&&= \frac{\cos \theta + e \cos^2 \theta+e \sin^2 \theta}{-\sin \theta} \\
&&&= \frac{\cos \theta + e}{-\sin \theta}
\end{align*}

Since our frame is rotated by $45^\circ$ we need to consider the appropriate gradient for this. 

We know that $m = \tan \theta$ so $m' = \tan (\theta+45^{\circ})  = \frac{1+m}{1-m}$

therefore we should have

\begin{align*}
&& \frac{\d y}{ \d x} &= \frac{1+\frac{\cos \theta + e}{-\sin \theta}}{1-\frac{\cos \theta + e}{-\sin \theta}} \\
&&&= \frac{\cos \theta - \sin \theta + e}{-\sin \theta - \cos \theta-e} \\
&&&= \frac{\sin \theta - \cos \theta -e}{\sin \theta + \cos \theta +e}
\end{align*}

As required.

The tangents at those points are parallel, therefore

\begin{align*}
&& \frac{\cos \alpha+e}{\sin \alpha} &= \frac{\cos \beta+e}{\sin \beta} \\
\Rightarrow && \frac{\frac{1-\tan^2 \frac{\alpha}{2}}{1+\tan^2 \frac{\alpha}{2}}+e}{\frac{2\tan \frac{\alpha}{2}}{1+\tan^2 \frac{\alpha}{2}}} &= \frac{\frac{1-\tan^2 \frac{\beta}{2}}{1+\tan^2 \frac{\beta}{2}}+e}{\frac{2\tan \frac{\beta}{2}}{1+\tan^2 \frac{\beta}{2}}} \\
&& \frac{1-\tan^2 \frac{\alpha}{2}+e(1+\tan^2\frac{\alpha}{2})}{2\tan\frac{\alpha}{2}} &=  \frac{1-\tan^2 \frac{\beta}{2}+e(1+\tan^2\frac{\beta}{2})}{2\tan\frac{\beta}{2}} \\
&& \frac{(1+e)+(e-1)\tan^2 \frac{\alpha}{2}}{2\tan \frac{\alpha}{2}} &= \frac{(1+e)+(e-1)\tan^2 \frac{\beta}{2}}{2\tan \frac{\beta}{2}} \\
&& \frac{(1+e)}{\tan\frac{\alpha}2} - (1-e)\tan\frac{\alpha}2 &=  \frac{(1+e)}{\tan\frac{\beta}2} - (1-e)\tan\frac{\beta}2
\end{align*}

ie both $\tan \frac{\alpha}{2}$ and $\tan \frac{\beta}{2}$ are roots of a quadratic of the form $(1-e)x^2-cx-(1+e)$ but this means the product of the roots is $-\frac{1+e}{1-e}$