The difference of the slopes of the lines rep | vedclass

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Question

(C) The given equation is of the form $ax^2 + 2hxy + by^2 = 0$,where $a = \sec^2 	heta - \sin^2 	heta$,$2h = -2 	an 	heta$,and $b = \sin^2 	heta$

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The given equation is of the form $ax^2 + 2hxy + by^2 = 0$,where $a = \sec^2 	heta - \sin^2 	heta$,$2h = -2 	an 	heta$,and $b = \sin^2 	heta$.Let $m_1$ and $m_2$ be the slopes of the lines. Then $m_1 + m_2 = -\frac{2h}{b} = \frac{2 	an 	heta}{\sin^2 	heta}$ and $m_1 m_2 = \frac{a}{b} = \frac{\sec^2 	heta - \sin^2 	heta}{\sin^2 	heta} = \sec^2 	heta \csc^2 	heta - 1$.The difference of slopes is $|m_1 - m_2| = \sqrt{(m_1 + m_2)^2 - 4m_1 m_2}$.$|m_1 - m_2| = \sqrt{\left(\frac{2 	an 	heta}{\sin^2 	heta}ight)^2 - 4(\sec^2 	heta \csc^2 	heta - 1)}$.$|m_1 - m_2| = \sqrt{\frac{4 	an^2 	heta}{\sin^4 	heta} - 4(\frac{1}{\cos^2 	heta \sin^2 	heta} - 1)}$.Since $\frac{	an^2 	heta}{\sin^4 	heta} = \frac{\sin^2 	heta}{\cos^2 	heta \sin^4 	heta} = \frac{1}{\cos^2 	heta \sin^2 	heta}$,the expression becomes $\sqrt{4(\frac{1}{\cos^2 	heta \sin^2 	heta} - \frac{1}{\cos^2 	heta \sin^2 	heta} + 1)} = \sqrt{4} = 2$.

The difference of the slopes of the lines represented by the equation ${x^2}(\sec^2 \theta - \sin^2 \theta) - 2xy \tan \theta + y^2 \sin^2 \theta = 0$ is:

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