MHT CET · Maths · Pair of Lines
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 \(\tan (\alpha+\beta)=\)
- A \(\frac{2 h}{b-a}\)
- B \(\frac{2 h}{a-b}\)
- C \(\frac{h}{a+b}\)
- D \(\frac{2 h}{a+b}\)
Answer & Solution
Correct Answer
(B) \(\frac{2 h}{a-b}\)
Step-by-step Solution
Detailed explanation
We have \(\tan \alpha+\tan \beta=\frac{-2 \mathrm{~h}}{\mathrm{~b}}\) and \(\tan \alpha \cdot \tan \beta=\frac{\mathrm{a}}{\mathrm{b}}\)
\(
\begin{aligned}
& \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} \\
& =\frac{\left(\frac{-2 h}{b}\right)}{1-\left(\frac{a}{b}\right)}=\frac{-2 h}{b} \times \frac{b}{b-a}=\frac{2 h}{a-b}
\end{aligned}
\)
\(
\begin{aligned}
& \tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} \\
& =\frac{\left(\frac{-2 h}{b}\right)}{1-\left(\frac{a}{b}\right)}=\frac{-2 h}{b} \times \frac{b}{b-a}=\frac{2 h}{a-b}
\end{aligned}
\)
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