Problem source

Show that, for a given constant $\gamma$ $(\sin\gamma\neq0)$ and with suitable choice of the constants $A$ and $B$, the line with cartesian equation $lx+my=1$ has polar equations 
\[
\frac{1}{r}=A\cos\theta+B\cos(\theta-\gamma).
\]
The distinct points $P$ and $Q$ on the conic with polar equations
\[
\frac{a}{r}=1+e\cos\theta
\]
correspond to $\theta=\gamma-\delta$ and $\theta=\gamma+\delta$ respectively, and $\cos\delta\neq0.$ Obtain the polar equation of the chord $PQ.$ Hence, or otherwise, obtain the equation of the tangent at the point where $\theta=\gamma.$
The tangents at $L$ and $M$ to a conic with focus $S$ meet at $T.$
Show that $ST$ bisects the angle $LSM$ and find the position of the intersection of $ST$ and $LM$ in terms of your chosen parameters for $L$ and $M.$

Solution source

\begin{align*}
&& \frac1{r} &= A \cos \theta + B \cos (\theta - \gamma) \\
&&&= A \cos \theta + B \cos \theta \cos \gamma + B  \sin \theta \sin \gamma \\
&&&= (A+B \cos \gamma) \cos \theta + B \sin \gamma \sin \theta \\
\Longleftrightarrow && 1 &= (A+B \cos \gamma) x + B \sin \gamma y
\end{align*}

So if we choose $B = \frac{m}{\sin \gamma}$ and $A = l-m \cot \gamma$ we have the desired result.

\begin{align*}
&& \frac{1 + e \cos (\gamma -\delta)}a &= A \cos (\gamma - \delta) + B \cos (\gamma - \delta - \gamma) \\
&&&= A \cos(\gamma-\delta) +B \cos \delta\\
&& \frac{1 + e \cos (\gamma +\delta)}{a}  &= A \cos (\gamma + \delta) + B \cos (\gamma + \delta - \gamma) \\
&&&= A \cos(\gamma + \delta) + B \cos \delta\\
\Rightarrow && \frac1{r} &= \frac{e}{a} \cos \theta + \frac{1}{a \cos \delta} \cos (\theta - \gamma) \\
\lim{\delta \to 0} &&\frac1{r} &= \frac{e}{a} \cos \theta+ \frac{1}{a} \cos (\theta - \gamma)
\end{align*}

Suppose we have have points $L$ and $M$ with $\theta = \gamma_L, \gamma_M$ then our tangents are:

\begin{align*}
&& \frac{a}{r} &= \cos \theta + \cos (\theta - \gamma_L) \\
&& \frac{a}{r} &= \cos \theta + \cos (\theta - \gamma_M) \\
\Rightarrow && 0 &= \cos (\theta - \gamma_L) -\cos(\theta - \gamma_M) \\
&&&= - 2 \sin \frac{(\theta - \gamma_L)+(\theta - \gamma_M)}{2} \sin \frac{(\theta - \gamma_L)-(\theta - \gamma_M)}{2} \\
&&&= -2 \sin \left ( \theta - \frac{\gamma_L+\gamma_M}2 \right) \sin \left ( \frac{\gamma_M - \gamma_L}{2}\right) \\
\Rightarrow && \theta &= \frac{\gamma_L+\gamma_M}{2}
\end{align*}

Therefore clearly $ST$ bisects $LSM$. 

The line $LM$ can be seen as the chord from the points $\frac{\gamma_L+\gamma_M}{2} \pm \frac{\gamma_L-\gamma_M}{2}$, so the line is:

\begin{align*}
&& \frac{a}{r} &= e \cos \theta + \frac{1}{\cos \left ( \frac{\gamma_L-\gamma_M}{2}\right)} \cos \left (\theta - \frac{\gamma_L+\gamma_M}{2} \right)
\end{align*}

and we want the point on the line where $\theta =\frac{\gamma_L+\gamma_M}{2}$ so 

\begin{align*}
&& \frac{a}{r} &= e \cos \left ( \frac{\gamma_L+\gamma_M}{2} \right) + \frac{1}{\cos \left ( \frac{\gamma_L-\gamma_M}{2}\right)} \\
\Rightarrow && r &= \frac{a}{e \cos \left ( \frac{\gamma_L+\gamma_M}{2} \right) + \frac{1}{\cos \left ( \frac{\gamma_L-\gamma_M}{2}\right)}}
\end{align*}