The angle theta between the pair of straight | vedclass

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(B) The given equation is a homogeneous second-degree equation $ax^2 + 2hxy + by^2 = 0$,which represents a pair of straight lines passing through the

Vedclass · Accepted Answer

$	an 	heta = \frac{2\sqrt{h^2 - ab}}{|a + b|}$

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(B) The given equation is a homogeneous second-degree equation $ax^2 + 2hxy + by^2 = 0$,which represents a pair of straight lines passing through the origin.Let the lines be $y = m_1x$ and $y = m_2x$.Then $m_1 + m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{a}{b}$.The angle $	heta$ between these lines is given by $	an 	heta = |\frac{m_1 - m_2}{1 + m_1m_2}|$.Using the identity $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1m_2$,we get:$|m_1 - m_2| = \sqrt{(m_1 + m_2)^2 - 4m_1m_2} = \sqrt{\frac{4h^2}{b^2} - \frac{4a}{b}} = \frac{2\sqrt{h^2 - ab}}{|b|}$.Substituting these into the formula for $	an 	heta$:$	an 	heta = |\frac{\frac{2\sqrt{h^2 - ab}}{b}}{1 + \frac{a}{b}}| = |\frac{2\sqrt{h^2 - ab}}{a + b}|$.Thus,the correct option is $B$.

The angle $\theta$ between the pair of straight lines represented by the homogeneous equation $ax^2 + 2hxy + by^2 = 0$ is given by:

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