Question

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

• A. $2{\tan }^{-1}\left(\frac{b}{a}\right)$
• B. ${\tan }^{-1}\left(\frac{2ab}{a^2-b^2}\right)$
• C. ${\tan }^{-1}\left(\frac{a^2-b^2}{a^2+b^2}\right)$
• D. $Noneofthese$

Answer 1

The correct option is B

${\tan }^{-1}\left(\frac{2ab}{a^2-b^2}\right)$

Finding the angle :

Given,

$\frac{x}{a}+\frac{y}{b}=1...\left(i\right)\&\frac{x}{a}-\frac{y}{b}=1...\left(ii\right)$

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

$y=-\frac{b}{a}x+1$ where slope $m_1=-\frac{b}{a}$

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

$y=\frac{b}{a}x+1$where slope $m_2=\frac{b}{a}$

Now angle between line $\left(i\right)$ & $\left(ii\right)$ is

$\begin{array}{rcl}tan\theta & =& \left|\frac{\mathrm{m}1+\mathrm{m}2}{1-\mathrm{m}1.\mathrm{m}2}\right|\\ & =& \left|\frac{{\displaystyle \frac{-b}{a}}-{\displaystyle \frac{b}{a}}}{1-{\displaystyle \frac{b^2}{a^2}}}\right|\\ \tan \theta & =& \left|\frac{-{\displaystyle \frac{2b}{a}}}{{\displaystyle \frac{a^2-b^2}{a^2}}}\right|\\ & =& \left|\frac{-2ab}{a^2-b^2}\right|\\ & =& \left|\frac{-2ab}{-b^2+a^2}\right|\\ \theta & =& {\tan }^{-1}\left|\frac{-2ab}{-b^2+a^2}\right|\end{array}$

Therefore, angle between $\left(i\right)$ & $\left(ii\right)$ is $\begin{array}{rcl}\theta & =& {\tan }^{-1}\left|\frac{-2ab}{-b^2+a^2}\right|\end{array}$

Hence, correct option is $\left(B\right)$