Find the difference between the slopes of the lines given by \((\tan^2\theta+\cos^2\theta)x^2-2xy\tan\theta+(\sin^2\theta)y^2=0\)

Answer: The difference between the slopes is 2.

Comparing with \(ax^2+2hxy+by^2=0\): \(a=\tan^2\theta+\cos^2\theta\), \(h=-\tan\theta\), \(b=\sin^2\theta\).

If \(m_1,m_2\) are the slopes, \(m_1+m_2=-\dfrac{2h}{b}=\dfrac{2\tan\theta}{\sin^2\theta}\) and \(m_1m_2=\dfrac ab=\dfrac{\tan^2\theta+\cos^2\theta}{\sin^2\theta}\).

\((m_1-m_2)^2=(m_1+m_2)^2-4m_1m_2=\dfrac{4\tan^2\theta}{\sin^4\theta}-\dfrac{4(\tan^2\theta+\cos^2\theta)}{\sin^2\theta}\)

\(=\dfrac{4}{\sin^4\theta}\left[\tan^2\theta-\sin^2\theta\tan^2\theta-\sin^2\theta\cos^2\theta\right]=\dfrac{4}{\sin^4\theta}\left[\tan^2\theta\cos^2\theta-\sin^2\theta\cos^2\theta\right]\)

\(=\dfrac{4}{\sin^4\theta}\left[\sin^2\theta(1-\cos^2\theta)\right]=\dfrac{4\sin^4\theta}{\sin^4\theta}=4\).

\(\therefore |m_1-m_2|=2\).