If two lines represented by the equation x^2 | vedclass

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Question

(C) The given equation is $x^2 - (1 + \sqrt{3})xy + \sqrt{3}y^2 = 0$.This is a homogeneous equation of the form $ax^2 + 2hxy + by^2 = 0$, where $a = 1

Vedclass · Accepted Answer

$\frac{\sqrt{3} + 1}{\sqrt{3} - 1}$

Vedclass · Answer

(C) The given equation is $x^2 - (1 + \sqrt{3})xy + \sqrt{3}y^2 = 0$.This is a homogeneous equation of the form $ax^2 + 2hxy + by^2 = 0$, where $a = 1$, $2h = -(1 + \sqrt{3})$, and $b = \sqrt{3}$.Let $m_1 = 	an \alpha$ and $m_2 = 	an \beta$ be the slopes of the two lines.From the properties of homogeneous equations, $m_1 + m_2 = -\frac{2h}{b} = \frac{1 + \sqrt{3}}{\sqrt{3}}$ and $m_1 m_2 = \frac{a}{b} = \frac{1}{\sqrt{3}}$.We need to find $	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta} = \frac{m_1 + m_2}{1 - m_1 m_2}$.Substituting the values: $	an(\alpha + \beta) = \frac{\frac{1 + \sqrt{3}}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \frac{\frac{1 + \sqrt{3}}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{1 + \sqrt{3}}{\sqrt{3} - 1}$.Rationalizing the denominator: $\frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{3 + 1 + 2\sqrt{3}}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$.Since the options provided are in the form of fractions, we observe that $\frac{1 + \sqrt{3}}{1 - \sqrt{3}} = \frac{(1 + \sqrt{3})^2}{1 - 3} = \frac{1 + 3 + 2\sqrt{3}}{-2} = -(2 + \sqrt{3})$.Re-evaluating the expression $\frac{1 + \sqrt{3}}{\sqrt{3} - 1}$, it matches option $C$.

If two lines represented by the equation $x^2 - (1 + \sqrt{3})xy + \sqrt{3}y^2 = 0$ make angles $\alpha$ and $\beta$ with the $X$-axis, then find the value of $\tan(\alpha + \beta)$.

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