The angle between the lines $a b\left(x^2-y^2\right)+\left(a^2-b^2\right) x y=0$ is
The angle between the lines $a b\left(x^2-y^2\right)+\left(a^2-b^2\right) x y=0$ is
$\frac{\pi}{2}$
$\frac{\pi}{3}$
$\frac{\pi}{4}$
$\frac{\pi}{6}$
Solution
We have,
$
\begin{aligned}
& a b\left(x^2-y^2\right)+\left(a^2-b^2\right) x y=0 \\
& a b x^2+\left(a^2-b^2\right) x y-a b y^2=0
\end{aligned}
$
Sum of the coefficient of $x^2$ and $y^2$
$
\text { i.e. } a b-a b=0
$
$\therefore \quad$ Lines are perpendicular to each other.
$\therefore \quad$ Angle between lines $=\pi / 2$