Question

Two conic sections have a common focus about which one of them is turned ; prove that the common chord is always a tangent to another conic, having the same focus, and whose eccentricity is the ratio of the eccentricities of the given conics.

Answer 1

Let the equation to the fixed conic be
$\frac{l}{r}=1-e\cos \theta$ and of the variable conic be $\frac{l_1}{r}=1-e_1\cos (\theta -\alpha )$, $\alpha$ having variable,
The equation to the common chord may be, obtained by subtracting the two, i.e. the common chord will be
$\frac{l}{r}-\frac{l_1}{r}=e_1\cos (\theta -\alpha )-e\cos \theta$
or $\frac{l-l_1}{r}=e_1[\cos (\theta -\alpha )-\frac{e}{e_1}\cos \theta ]$
or $\frac{\frac{1}{e_1}(l-l_1)}{r}=\cos (\theta -\alpha )-\frac{e}{e_1}\cos \theta$
Clearly this line touches the conic
$\frac{\frac{1}{e_1}(l-l_1)}{r}=1-\frac{e}{e_1}\cos \theta$
Similarly, we can prove that the other common chord,
$\frac{\frac{1}{e_1}(l-l_1)}{r}=\cos (\theta -\alpha )+\frac{e}{e_1}\cos \theta$
touch the conic
$\frac{\frac{1}{e_1}(l-l_1)}{r}=1+\frac{e}{e_1}\cos \theta$