Question

Eliminate $\theta$ from the relations $\alpha \sec \theta =1-b\tan \theta$
and ${\alpha }^2\sec {}^2\theta =5+b^2{\tan }^2\theta$

Answer 1

Squareing first relation and putting in $2nd$,
$1-2b\tan \theta +b^2{\tan }^2\theta =5+b^2{\tan }^2\theta$
which gives $\theta =-2/b$.
Putting this value of $\tan \theta$ in the second given relation, we get
$a^2[1+(4/b^2\left)\right]=5+b^2.\left(4/b^2\right)=9$
or $a^2{b}^2+4a^2=9b^2$.