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If the angle between the lines represented by the equation $x^2+\lambda x y-y^2 \tan ^2 \theta=0$ is $2 \theta$, then the value of $\lambda$ is
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The correct answer is:
$0$
Given equation of pair of lines is
$$
\begin{array}{ll}
& x^2+\lambda x y-y^2 \tan ^2 \theta=0 \\
\therefore \quad & \mathrm{a}=1, \mathrm{~h}=\frac{\lambda}{2}, \mathrm{~b}=-\tan ^2 \theta \\
\therefore \quad & \tan 2 \theta=\left|\frac{2 \sqrt{\mathrm{h}^2-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\right| \\
\Rightarrow & \frac{2 \tan \theta}{1-\tan ^2 \theta}=\left|\frac{2 \sqrt{\frac{\lambda^2}{4}+\tan ^2 \theta}}{1-\tan ^2 \theta}\right| \\
\Rightarrow & \frac{\lambda^2}{4}+\tan ^2 \theta=\tan ^2 \theta \\
\Rightarrow & \lambda=0
\end{array}
$$
$$
\begin{array}{ll}
& x^2+\lambda x y-y^2 \tan ^2 \theta=0 \\
\therefore \quad & \mathrm{a}=1, \mathrm{~h}=\frac{\lambda}{2}, \mathrm{~b}=-\tan ^2 \theta \\
\therefore \quad & \tan 2 \theta=\left|\frac{2 \sqrt{\mathrm{h}^2-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\right| \\
\Rightarrow & \frac{2 \tan \theta}{1-\tan ^2 \theta}=\left|\frac{2 \sqrt{\frac{\lambda^2}{4}+\tan ^2 \theta}}{1-\tan ^2 \theta}\right| \\
\Rightarrow & \frac{\lambda^2}{4}+\tan ^2 \theta=\tan ^2 \theta \\
\Rightarrow & \lambda=0
\end{array}
$$
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