The square of the difference of the slopes of | vedclass

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Question

(C) The given equation is $x^2(\sec^2 	heta - \sin^2 	heta) - 2xy 	an 	heta + y^2 \sin^2 	heta = 0$. Comparing this with the general equation $ax

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) The given equation is $x^2(\sec^2 	heta - \sin^2 	heta) - 2xy 	an 	heta + y^2 \sin^2 	heta = 0$. Comparing this with the general equation $ax^2 + 2hxy + by^2 = 0$,we have: $a = \sec^2 	heta - \sin^2 	heta$,$h = -	an 	heta$,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 square of the difference of the slopes is $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2$. $(m_1 - m_2)^2 = \left(\frac{2 	an 	heta}{\sin^2 	heta}ight)^2 - 4\left(\frac{\sec^2 	heta - \sin^2 	heta}{\sin^2 	heta}ight)$. $(m_1 - m_2)^2 = \frac{4 \sin^2 	heta}{\cos^2 	heta \sin^4 	heta} - \frac{4(\sec^2 	heta - \sin^2 	heta)}{\sin^2 	heta} = \frac{4}{\cos^2 	heta \sin^2 	heta} - \frac{4}{\cos^2 	heta \sin^2 	heta} + 4 = 4$. Thus,the square of the difference of the slopes is $4$.

The square of 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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