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
\[
\tan(\alpha/2)\tan(\beta/2)=-\frac{1+e}{1-e}.
\]
[Hint. A diameter of an ellipse is a chord through its centre.]
Solution:
\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 \(-\frac{1+e}{1-e}\)
