Q79 / 180 · KCET 2013 · Mathematics
The angle between the lines $\sin ^{2} \alpha \cdot y^{2}-2 x y \cdot \cos ^{2} \alpha+\left(\cos ^{2} \alpha-1\right) x^{2}=0$, is
- Correct answerA$90^{\circ}$
- B$\alpha$
- C$\frac{\alpha}{2}$
- D$2 \alpha$
Solution
Given pair of straight line is
$\left(\cos ^{2} \alpha-1\right) x^{2}-2 \cos ^{2} \alpha \cdot x y+\sin ^{2} \alpha y^{2}=0$
On comparing with the equation
$\begin{aligned}
&a x^{2}+2 h x y+b y^{2}=0, \text { we get } \\
&a=\cos ^{2} \alpha-1=-\sin ^{2} \alpha \\
&h=-\cos ^{2} \alpha \text { and } b=\sin ^{2} \alpha
\end{aligned}$
Let $\theta$ be the angle between the lines, then
$\begin{aligned}
\tan \theta &=\left|\frac{2 \sqrt{h^{2}-a b}}{a+b}\right| \\
\Rightarrow \quad \tan \theta &=\left|\frac{2 \sqrt{\cos ^{4} \alpha+\sin ^{4} \alpha}}{-\sin ^{2} \alpha+\sin ^{2} \alpha}\right|=\infty=\tan 90^{\circ} \\
\Rightarrow \quad \theta &=90^{\circ}
\end{aligned}$
Hence, the correct answer is option: 'A'.