Question

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

• A. $ta{n}^{-1}\frac{a^2+b^2}{a^2-b^2}$
• B. $ta{n}^{-1}\frac{2ab}{a^2+b^2}$
• C. $ta{n}^{-1}\frac{a^2-b^2}{a^2+b^2}$
• D. $ta{n}^{-1}\frac{2ab}{a^2-b^2}$

Answer 1

Answer: (D)

$ta{n}^{-1}\frac{2ab}{a^2-b^2}$

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

$\Rightarrow y=b-\frac{b}{a}x,y=\frac{b}{a}x-b$

Slope $m_1=\frac{-b}{a},m_2=\frac{b}{a}$ [comparing with $y=mx+c$]

Angle between lens $={\tan }^{-1}\left|\frac{m_1+m_2}{1+m_1{m}_2}\right|$

$={\tan }^{-1}|left\left(\frac{-\frac{b}{a}-\frac{b}{a}}{1+(-\frac{b}{a})\left(\frac{b}{a}\right)}|right\right)={\tan }^{-1}|left\left(\frac{\alpha ba}{a^2-b^2}|right\right)$