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 $x$, then if $\theta =\frac{\pi }{6},\tan {\theta }_1+\tan {\theta }_2$ is equal to

• A. $(-\frac{8}{3})si{n}^2\alpha$
• B. $(-\frac{8}{3})cosec2\alpha$
• C. $-8\sqrt{3}cosec2\alpha$
• D. $-4cosec2\alpha$

Answer 1

The correct option is B $(-\frac{8}{3})cosec2\alpha$

The given equation can be written as
$(x^2+y^2)({\cos }^2\theta {\sin }^2\alpha +{\sin }^2\theta )=x^2{\tan }^2\alpha -2xy\tan \alpha \sin \theta +y^2{\sin }^2\theta$ or $({\cos }^2\theta {\sin }^2\alpha +{\sin }^2\theta -{\tan }^2\alpha )x^2+2\left(\tan \alpha \sin \theta \right)xy+{\cos }^2\theta {\sin }^2\alpha y^2=0$
Since the slope of these lines are given as $\tan {\theta }_1$ and $\tan {\theta }_2$.
and sum of the slopes $=\frac{-2\tan \alpha \sin \theta }{{\cos }^2\theta {\sin }^2\alpha }$
$\Rightarrow \tan {\theta }_1+\tan {\theta }_2=\frac{-2\tan \alpha }{2\times \left(\frac{3}{4}\right)\times {\sin }^2\alpha }(\because \theta =\frac{\pi }{6})=-\left(\frac{8}{3}\right)cosec2\alpha$