If two lines represented by $a x^2+2 h x y+b^2=0$ makes angles $\alpha$ and $\beta$ with positive direction…
If two lines represented by $a x^2+2 h x y+b^2=0$ makes angles $\alpha$ and $\beta$ with positive direction of $\mathrm{X}$-axis, then $\tan (\alpha+\beta)=$
$\frac{2 h}{b-a}$
$\frac{2 h}{a-b}$
$\frac{h}{a+b}$
$\frac{2 h}{a+b}$
Solution
We have $\tan \alpha+\tan \beta=\frac{-2 \mathrm{~h}}{\mathrm{~b}}$ and $\tan \alpha \cdot \tan \beta=\frac{\mathrm{a}}{\mathrm{b}}$
$\begin{aligned}
& \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} \\
& =\frac{\left(\frac{-2 h}{b}\right)}{1-\left(\frac{a}{b}\right)}=\frac{-2 h}{b} \times \frac{b}{b-a}=\frac{2 h}{a-b}
\end{aligned}$