The angle between the lines frac{x}{a} + frac | vedclass

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Question

(A) The given lines are $\frac{x}{a} + \frac{y}{b} = 1$ and $\frac{x}{a} - \frac{y}{b} = 1$.These can be rewritten as $bx + ay - ab = 0$ and $bx - ay

Vedclass · Accepted Answer

$2	an^{-1}\left(\frac{b}{a}ight)$

Vedclass · Answer

(A) The given lines are $\frac{x}{a} + \frac{y}{b} = 1$ and $\frac{x}{a} - \frac{y}{b} = 1$.These can be rewritten as $bx + ay - ab = 0$ and $bx - ay - ab = 0$.The slopes of these lines are $m_1 = -\frac{b}{a}$ and $m_2 = \frac{b}{a}$.The angle $	heta$ between two lines with slopes $m_1$ and $m_2$ is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$.Substituting the values,$	an 	heta = \left| \frac{-\frac{b}{a} - \frac{b}{a}}{1 + (-\frac{b}{a})(\frac{b}{a})} ight| = \left| \frac{-\frac{2b}{a}}{1 - \frac{b^2}{a^2}} ight| = \left| \frac{-\frac{2b}{a}}{\frac{a^2 - b^2}{a^2}} ight| = \left| \frac{2ab}{a^2 - b^2} ight|$.Using the identity $2	an^{-1}(x) = 	an^{-1}\left(\frac{2x}{1-x^2}ight)$,we have $2	an^{-1}\left(\frac{b}{a}ight) = 	an^{-1}\left(\frac{2(b/a)}{1-(b/a)^2}ight) = 	an^{-1}\left(\frac{2ab}{a^2-b^2}ight)$.Thus,the angle is $2	an^{-1}\left(\frac{b}{a}ight)$.

The angle between the lines $\frac{x}{a} + \frac{y}{b} = 1$ and $\frac{x}{a} - \frac{y}{b} = 1$ is

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