MHT CET · Maths · Pair of Lines
If the lines represented by \(a x^2-b x y-y^2=0\) make angle \(\alpha\) and \(\beta\) with the positive direction of \(X\)-axis, then \(\tan (\alpha+\beta)=\)
- A \(\frac{a}{a+b}\)
- B \(\frac{\mathrm{b}}{1+\mathrm{b}}\)
- C \(\frac{\mathrm{b}}{1+\mathrm{a}}\)
- D \(\frac{-b}{1+a}\)
Answer & Solution
Correct Answer
(D) \(\frac{-b}{1+a}\)
Step-by-step Solution
Detailed explanation
\(\tan \alpha\) and \(\tan \beta\) are roots of the \(a x^2-b x y-y^2=0\)
\(\therefore \tan \alpha+\tan \beta=\frac{-(- b )}{-1}=- b\) and \(\tan \alpha \tan \beta=\frac{ a }{(-1)}=- a\)
\(\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\frac{-\mathrm{b}}{1-(-\mathrm{a})}=\frac{-\mathrm{b}}{1+\mathrm{a}}\)
\(\therefore \tan \alpha+\tan \beta=\frac{-(- b )}{-1}=- b\) and \(\tan \alpha \tan \beta=\frac{ a }{(-1)}=- a\)
\(\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\frac{-\mathrm{b}}{1-(-\mathrm{a})}=\frac{-\mathrm{b}}{1+\mathrm{a}}\)
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