Q89 / 160 · AP EAMCET 2004 · Mathematics
The angle between the lines represented by $y^2 \sin ^2 \theta-x y \sin ^2 \theta+x^2\left(\cos ^2 \theta-1\right)=0$ is
- A$\frac{\pi}{3}$
- B$\frac{\pi}{4}$
- C$\frac{\pi}{6}$
- Correct answerD$\frac{\pi}{2}$
Solution
Equation of lines is
$$
x^2\left(\cos ^2 \theta-1\right)-x y \sin ^2 \theta+y^2 \sin ^2 \theta=0
$$
This is a homogeneous equation of second degree, on comparing with $a x^2+2 h x y+b y^2=0$
$$
\Rightarrow a=\cos ^2 \theta-1, \mathrm{~b}=\sin ^2 \theta, h=-\frac{1}{2} \sin ^2 \theta
$$
Now,
$$
\begin{aligned}
a+b & =\cos ^2 \theta-1+\sin ^2 \theta \\
& =1-1=0
\end{aligned}
$$
$\begin{array}{ll}\therefore & \tan \theta=\frac{2 \sqrt{h^2-a b}}{a+b} \\ \Rightarrow & \tan \theta=\infty \Rightarrow \theta=\frac{\pi}{2}\end{array}$
Hence, the correct answer is option: 'D'.