Question

The difference of the tangents of the angles which the lines $\left({\tan }^2\alpha +{\cos }^2\alpha \right)x^2-2xy\tan \alpha +\left({\sin }^2\alpha \right)y^2=0$ make with the $X-$ axis is

• A. $\frac{1}{2}$
• B. $1$
• C. $2$
• D. $\frac{\sqrt{3}-1}{2}$

Answer 1

Answer: (C)

We have
$x^2\left({\tan }^2\alpha +{\cos }^2\alpha \right)-2xy\tan \alpha +y^2{\sin }^2\alpha =0$
Comparing this equation with $a{x}^2+2hxy+b{y}^2=0$,
We have,
$a={\tan }^2\alpha +{\cos }^2\alpha ,2h=-2\tan \alpha$ and $b={\sin }^2\alpha$
Let $m_1$ and $m_2$ be the slopes of the lines represented by the given equation, then
$m_1+m_2=\frac{2\tan \alpha }{{\sin }^2\alpha }=\frac{2}{\sin \alpha \cos \alpha }$
and $m_1{m}_2=\frac{{\tan }^2\alpha +{\cos }^2\alpha }{{\sin }^2\alpha }=\frac{1}{{\cos }^2\alpha }+\frac{{\cos }^2\alpha }{{\sin }^2\alpha }$ $\therefore m_1-m_2=\sqrt{{\left(m_1+m_2\right)}^2-4m_1{m}_2}$ $=\sqrt{\frac{4}{{\sin }^2\alpha \cdot {\cos }^2\alpha }-4\frac{{\cos }^2\alpha +{\tan }^2\alpha }{{\sin }^2\alpha }}$
$=\sqrt{\frac{4-4{\cos }^2\alpha \left({\cos }^2\alpha +{\tan }^2\alpha \right)}{{\sin }^2\alpha {\cos }^2\alpha }}$
$=\sqrt{\frac{4-4{\cos }^4\alpha -4{\sin }^2\alpha }{{\sin }^2\alpha {\cos }^2\alpha }}$
$=\sqrt{\frac{4{\cos }^2\alpha -4{\cos }^4\alpha }{{\sin }^2\alpha {\cos }^2\alpha }}$
$=\sqrt{\frac{4{\cos }^2\alpha \left(1-{\cos }^2\alpha \right)}{{\cos }^2\alpha {\sin }^2\alpha }}$
$=\sqrt{\frac{4{\cos }^2\alpha \cdot {\sin }^2\alpha }{{\cos }^2\alpha \cdot {\sin }^2\alpha }}=\sqrt{4}=2$