Question

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

Answer 1

We have
$$\begin{gathered} \frac{x}{a}+\frac{y}{b}=1\mathrm{and}\frac{x}{a}-\frac{y}{b}=1 \\ \Rightarrow \frac{y}{b}=-\frac{x}{a}+1\mathrm{and}\frac{y}{b}=\frac{x}{a}-1 \\ \Rightarrow y=-\frac{b}{a}x+b\mathrm{and}y=\frac{b}{a}-b \end{gathered}$$
The slopes of the two lines are $-\frac{b}{a}\mathrm{and}\frac{b}{a}$
Now, the tangent of an angle between the lines is given by
$$\begin{gathered} \frac{{\displaystyle \frac{b}{a}}+{\displaystyle \frac{b}{a}}}{1-{\displaystyle \frac{b}{a}}\times {\displaystyle \frac{b}{a}}} \\ =\frac{{\displaystyle \frac{2b}{a}}}{{\displaystyle \frac{a^2-b^2}{a^2}}} \\ =\frac{2ab}{a^2-b^2} \end{gathered}$$