Browse Questions

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

Toolbox:
• Angle between two lines is $\theta = \tan^{-1} \bigg| \large\frac{m_1-m_2}{1+m_1m_2} \bigg|$ where $m_1$ and $m_2$ are the slopes of the two lines.
Step 1:
the equation of the given lines are :
$\large\frac{x}{a}$$+\large\frac{y}{b}$$=1$-----(1)
$\large\frac{x}{a}$$-\large\frac{y}{b}$$=1$--------(2)
This can be written as
$bx+ay=ab$
or $y=-\large\frac{bx}{a}$$+b----------(3) bx-ay=ab or y = \large\frac{b}{a}$$x-b$--------(4)
Hence the slopes of the lines (1) and (2) are $-\large\frac{b}{a}$ and $\large\frac{b}{a}$ respectively.
Step 2 :
The angle between the lines is
$\theta = \tan^{-1} \bigg| \large\frac{m_1-m_2}{1+m_1m_2} \bigg|$
Substituting for $m_1$ and $m_2$ we get,
$\theta = \tan^{-1} \bigg| \large\frac{ -\Large\frac{b}{a}-\Large\frac{b}{a}}{1+\Large\frac{b}{a} \times \large\frac{b}{a}} \bigg|$
$= \tan^{-1} \bigg| \large\frac{-2\Large\frac{b}{a}}{\Large\frac{a^2+b^2}{a^2}} \bigg|$
$\therefore \tan \theta = \large\frac{2ab}{a^2+b^2}$
Hence proved.