Here's what I have done so far (in brief):
Let us assume a conic to be $$\frac{l}{r}=1+e\cos{\theta}\tag{a}$$ in the Polar Coordinate space with the focus of the conic as the Pole. Let us assume two points on the conic $P$ and $Q$ on the conic with coordinates $(r_1, \alpha-\beta)$ and $(r_2, \alpha+\beta)$. Now, considering a line passing through $P$ and $Q$ with the equation $$\frac{l}{r}=A\cos{\theta}+B\sin{\theta},\tag{b}$$ we must have $P$ and $Q$ satisfy both the equations of the conic and the straight line.
For the point $P$, we have:
$$\frac{l}{r_1}=1+e\cos{(\alpha-\beta)}=A\cos{(\alpha-\beta)}+B\sin{(\alpha-\beta)}\tag{1}$$
For the point $Q$, we have:
$$\frac{l}{r_2}=1+e\cos{(\alpha+\beta)}=A\cos{(\alpha+\beta)}+B\sin{(\alpha+\beta)}\tag{2}$$
Multiplying $(1)$ by $\cos{(\alpha+\beta)}$ and $(2)$ by $\cos{(\alpha-\beta)}$, then subtracting and simplifying, I arrived at
$$B=\frac{\sin{\alpha}}{\cos{\beta}}$$
However, if I try to plug this value into the original equation, i.e. $(a)$, the result is very messy and it quickly becomes unworkable (for me). How do I proceed (in a simple manner) to find $A$ and finally, the equation of the chord?
Best Answer
Continuing with the same method adopted earlier, we multiply equations $(1)$ and $(2)$ with $\sin(\alpha+\beta)$ and $\sin(\alpha-\beta)$ respectively. Then we subtract the equations, rearrange and simplify to get $$A-e=\frac{\cos{\alpha}}{\cos{\beta}}\tag{3}$$
We then use this value to obtain the general equation of the chord as $$\begin{align}\frac{l}{r} &= \left(e+\frac{\cos{\alpha}}{\cos{\beta}}\right)\cdot\cos{\theta}+\frac{\sin{\alpha}}{\cos{\beta}}\cdot\sin{\theta} \\ &= e\cos{\theta}+\frac{1}{\cos{\beta}}\cdot\left(cos{\alpha}\cdot\cos{\theta}+sin{\alpha}\cdot\sin{\theta }\right) \\ &= e\cos{\theta}+\frac{\cos{(\alpha-\theta)}}{\cos{\beta}} \\ &= e\cos{\theta}+\frac{\cos{(\theta-\alpha)}}{\cos{\beta}} \tag{4} \end{align}$$
We can go one step further and obtain the equation of a tangent to the conic. For this, we need to set $\beta=0$ which would imply $\cos{\beta}=1$. Thus, the equation of the tangent would be: $$\frac{l}{r}=e\cos{\theta}+\cos{(\theta-\alpha)}$$