MHT CET · Maths · Pair of Lines
If the two lines given by \(a x^2+2 \mathrm{hxy}+\mathrm{by}^2=0\) make inclinations \(\alpha\) and \(\beta\), then \(\tan (\alpha+\beta)=\)
- A \(\frac{\mathrm{h}}{\mathrm{a}+\mathrm{b}}\)
- B \(\frac{2 h}{a+b}\)
- C \(\frac{h}{a-b}\)
- D \(\frac{2 h}{a-b}\)
Answer & Solution
Correct Answer
(D) \(\frac{2 h}{a-b}\)
Step-by-step Solution
Detailed explanation
Lines given by ax \(2+2 h x y+b^2=0\) make inclinations \(\alpha\) and \(\beta\).
\[
\therefore \tan \alpha+\tan \beta=\frac{-2 \mathrm{~h}}{\mathrm{~b}} \text { and } \tan \alpha \tan \beta=\frac{\mathrm{a}}{\mathrm{b}}
\]
Now \(\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\frac{\left(-\frac{2 h}{b}\right)}{1-\left(\frac{a}{b}\right)}=\frac{-2 h}{b} \times \frac{b}{(b-a)}\)
\[
\tan (\alpha+\beta)=\frac{-2 h}{b-a}=\frac{2 h}{a-b}
\]
\[
\therefore \tan \alpha+\tan \beta=\frac{-2 \mathrm{~h}}{\mathrm{~b}} \text { and } \tan \alpha \tan \beta=\frac{\mathrm{a}}{\mathrm{b}}
\]
Now \(\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\frac{\left(-\frac{2 h}{b}\right)}{1-\left(\frac{a}{b}\right)}=\frac{-2 h}{b} \times \frac{b}{(b-a)}\)
\[
\tan (\alpha+\beta)=\frac{-2 h}{b-a}=\frac{2 h}{a-b}
\]
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