Question

Show that the tangent of an angle between the lines $\frac{x}{a}+\frac{y}{b}=1and\frac{x}{a}-\frac{y}{b}=1is\frac{2ab}{a^2-b^2}$

Answer 1

Given $\frac{x}{a}+\frac{y}{b}=1$ …………$\left(1\right)$

$\frac{x}{a}-\frac{y}{b}=1$ ………….$\left(2\right)$

Equation $\left(1\right)$ can be written as

$y=-\frac{b}{a}x+1$

where slope $[m_1=\frac{-b}{a}]$

Equation $\left(2\right)$ can be written as

$y=\frac{b}{a}x-1$

where slope $[m_2=\frac{b}{a}]$

now angle between line $\left(1\right)$ & $\left(2\right)$ is

$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\left|\frac{\frac{-b}{a}-\frac{b}{a}}{1-\frac{b^2}{a^2}}\right|$

$\tan \theta =\left|\frac{-\frac{2b}{a}}{\frac{a^2-b^2}{a^2}}\right|=\left|\frac{-2ba}{a^2-b^2}\right|=\left|\frac{-2ab}{-b^2+a^2}\right|$

Hence tangent of an angle between line $\left(1\right)$ & $\left(2\right)$

is $\frac{2ab}{a^2-b^2}$.