Search any question & find its solution
Question:
Answered & Verified by Expert
Show that the tangent of an angle between the lines $\frac{x}{a}+\frac{y}{b}=1$ and $\frac{x}{a}-\frac{y}{b}=1$ is $\frac{2 a b}{a^2-b^2}$.
Solution:
2463 Upvotes
Verified Answer
Lines are: $\frac{x}{a}+\frac{y}{b}=1$ and $\frac{x}{a}-\frac{y}{b}=1$
$$
\begin{aligned}
&\therefore \mathrm{m}_1=\frac{-\mathrm{b}}{\mathrm{a}}, \mathrm{m}_2=\frac{\mathrm{b}}{\mathrm{a}} \\
&\therefore \quad \tan \theta=\left|\frac{-\frac{\mathrm{b}}{\mathrm{a}}-\frac{\mathrm{b}}{\mathrm{a}}}{1+\left(\frac{-\mathrm{b}}{\mathrm{a}}\right)\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}\right| \\
&\Rightarrow \tan \theta\left|\frac{-2 \mathrm{ab}}{\mathrm{a}^2-\mathrm{b}^2}\right|=\left|\frac{2 \mathrm{ab}}{\mathrm{a}^2-\mathrm{b}^2}\right|=\pm\left(\frac{2 \mathrm{ab}}{\mathrm{a}^2-\mathrm{b}^2}\right)
\end{aligned}
$$
$$
\begin{aligned}
&\therefore \mathrm{m}_1=\frac{-\mathrm{b}}{\mathrm{a}}, \mathrm{m}_2=\frac{\mathrm{b}}{\mathrm{a}} \\
&\therefore \quad \tan \theta=\left|\frac{-\frac{\mathrm{b}}{\mathrm{a}}-\frac{\mathrm{b}}{\mathrm{a}}}{1+\left(\frac{-\mathrm{b}}{\mathrm{a}}\right)\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}\right| \\
&\Rightarrow \tan \theta\left|\frac{-2 \mathrm{ab}}{\mathrm{a}^2-\mathrm{b}^2}\right|=\left|\frac{2 \mathrm{ab}}{\mathrm{a}^2-\mathrm{b}^2}\right|=\pm\left(\frac{2 \mathrm{ab}}{\mathrm{a}^2-\mathrm{b}^2}\right)
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.