Question

Prove that the two conics $\frac{l_1}{r}=1-e\cos \theta$ and $\frac{l_2}{r}=1-e\cos (\theta -a)$ will touch one another, if $l_1^2(1-e_2^2)+l_2^2(1-e_1^2)+2l_1{l}_2{e}_1{e}_2\cos \alpha =2l_1{l}_2.$

Answer 1

If $\beta$ be the vectorial angle of the common point of contact (say P), then tangent at this point with respect to $1$ will be
$\frac{l_1}{r}=\cos (\theta -\beta )-e_1\cos \theta$
or $\frac{l_1}{r}=\cos \theta (\cos \beta -e_1)+\sin \theta \sin \beta .....3$
Again the tangent at P w.r.t. $2$ will be
$\frac{l_2}{r}=\cos (\theta -\beta )-e_2\cos (\theta -\alpha )$
or $\frac{l_2}{r}=\cos \theta (\cos \beta -e_2\cos \alpha )+\sin \theta (\sin \beta -e_2\sin \alpha )....4$
If $3$ and $4$ represent the same straight line, comparing the two we have
$\frac{\cos \beta -e_1}{\cos \beta -e_2\cos \alpha }=\frac{\sin \beta }{\sin \beta -e_2\sin \alpha }=\frac{l_1}{l_2}$
Hence $\cos \beta =\frac{l_1{e}_2\cos \alpha -l_2{e}_1}{l_1-l_2}$
and $\sin \beta =\frac{l_1{e}_2\sin \alpha }{l_1-l_2}$
To eliminate $\beta$ squaring and adding the two, we get the required condition as
${(l_1-l_2)}^2={(l_1{e}_2\cos \alpha -l_2{e}_1)}^2+l_1^2{e}_2^{2}si{n}^2\alpha$
or $l_1^2(1-e_2^2)+l_2^2(1-e_1^2)=2l_1{l}_2(1-e_1{e}_2\cos \alpha )$