Question

If ${\theta }_1$ and ${\theta }_2$ be the angles which the lines
$(x^2+y^2)({\cos }^2\theta {\sin }^2\alpha +{\sin }^2\theta )={(x\tan \alpha -y\sin \theta )}^2$ make with the axis of $X$, $\theta =\frac{\pi }{6}$, then $\tan {\theta }_1+\tan {\theta }_2=-\frac{A}{B}\csc 2\alpha$. Find the value of $A-B$

Answer 1

It is given that

$(x^2+y^2)({\cos }^2\theta {\sin }^2\alpha +{\sin }^2\theta )=(x\tan \alpha -y\sin \theta {)}^2$

$\Rightarrow (x^2+y^2)({\cos }^2\theta {\sin }^2\alpha +{\sin }^2\theta )=x^2{\tan }^2\alpha +y^2{\sin }^2\theta -2xy\tan \alpha \sin \theta$

$\Rightarrow \left({\cos }^2\theta {\sin }^2\alpha \right)y^2+2\left(\tan \alpha \sin \theta \right)xy+({\cos }^2\theta {\sin }^2\alpha +{\sin }^2\theta -{\tan }^2\alpha )x^2=0$

Now, sum of the slopes $=-\frac{coefficientofxy}{coefficientof{y}^2}$

$\Rightarrow \tan {\theta }_1+\tan {\theta }_2=-\frac{2\tan \alpha \sin \theta }{{\cos }^2\theta {\sin }^2\alpha }$

Putting $\theta =\pi /6$

We get, $\tan {\theta }_1+\tan {\theta }_2=-\frac{2\sin \alpha \times \frac{1}{2}}{\frac{3}{4}{\sin }^2\alpha \cos \alpha }$

$\Rightarrow \tan {\theta }_1+\tan {\theta }_2=-\frac{8}{3\times 2\sin \alpha \cos \alpha }$

$\Rightarrow \tan {\theta }_1+\tan {\theta }_2=-\frac{8}{3}\csc 2\alpha$

On comparing this equation with given equation, $\tan {\theta }_1+\tan {\theta }_2=-\frac{A}{B}\csc 2\alpha$

We get $A=8andB=3$

$\therefore A-B=5$