Question

The difference of the tangents of the angles which the lines $\left(\tan ^{2} \alpha+\cos ^{2} \alpha\right) x^{2}-2 x y \tan \alpha+\left(\sin ^{2} \alpha\right) y^{2}=0$ 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
$x^{2}\left(\tan ^{2} \alpha+\cos ^{2} \alpha\right)-2 x y \tan \alpha+y^{2} \sin ^{2} \alpha=0$
Comparing this equation with $a x^{2}+2 h x y+b y^{2}=0$,
We have,
$a=\tan ^{2} \alpha+\cos ^{2} \alpha, 2 h=-2 \tan \alpha$ 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 $m_{1} m_{2}=\frac{\tan ^{2} \alpha+\cos ^{2} \alpha}{\sin ^{2} \alpha}=\frac{1}{\cos ^{2} \alpha}+\frac{\cos ^{2} \alpha}{\sin ^{2} \alpha}$ $\therefore m_{1}-m_{2}=\sqrt{\left(m_{1}+m_{2}\right)^{2}-4 m_{1} m_{2}}$ $=\sqrt{\frac{4}{\sin ^{2} \alpha \cdot \cos ^{2} \alpha}-4 \frac{\cos ^{2} \alpha+\tan ^{2} \alpha}{\sin ^{2} \alpha}}$
$=\sqrt{\frac{4-4 \cos ^{2} \alpha\left(\cos ^{2} \alpha+\tan ^{2} \alpha\right)}{\sin ^{2} \alpha \cos ^{2} \alpha}}$
$=\sqrt{\frac{4-4 \cos ^{4} \alpha-4 \sin ^{2} \alpha}{\sin ^{2} \alpha \cos ^{2} \alpha}}$
$=\sqrt{\frac{4 \cos ^{2} \alpha-4 \cos ^{4} \alpha}{\sin ^{2} \alpha \cos ^{2} \alpha}}$
$=\sqrt{\frac{4 \cos ^{2} \alpha\left(1-\cos ^{2} \alpha\right)}{\cos ^{2} \alpha \sin ^{2} \alpha}}$
$=\sqrt{\frac{4 \cos ^{2} \alpha \cdot \sin ^{2} \alpha}{\cos ^{2} \alpha \cdot \sin ^{2} \alpha}}=\sqrt{4}=2$