MHT CET · Maths · Pair of Lines
If the angle between the lines represented by the equation \(x^2+\lambda x y-y^2 \tan ^2 \theta=0\) is \(2 \theta\), then the value of \(\lambda\) is
- A \(0\)
- B \(1\)
- C \(\tan \theta\)
- D \(2\)
Answer & Solution
Correct Answer
(A) \(0\)
Step-by-step Solution
Detailed explanation
Given equation of pair of lines is
\(x^2+\lambda x y-y^2 \tan ^2 \theta=0 \)
\( \therefore \mathrm{a}=1, \mathrm{~h}=\frac{\lambda}{2}, \mathrm{~b}=-\tan ^2 \theta \)
\( \therefore \tan 2 \theta=\left|\frac{2 \sqrt{\mathrm{h}^2-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\right| \)
\( \Rightarrow \frac{2 \tan \theta}{1-\tan ^2 \theta}=\left|\frac{2 \sqrt{\frac{\lambda^2}{4}+\tan ^2 \theta}}{1-\tan ^2 \theta}\right| \)
\( \Rightarrow \frac{\lambda^2}{4}+\tan ^2 \theta=\tan ^2 \theta \)
\( \Rightarrow \lambda=0\)
\(x^2+\lambda x y-y^2 \tan ^2 \theta=0 \)
\( \therefore \mathrm{a}=1, \mathrm{~h}=\frac{\lambda}{2}, \mathrm{~b}=-\tan ^2 \theta \)
\( \therefore \tan 2 \theta=\left|\frac{2 \sqrt{\mathrm{h}^2-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\right| \)
\( \Rightarrow \frac{2 \tan \theta}{1-\tan ^2 \theta}=\left|\frac{2 \sqrt{\frac{\lambda^2}{4}+\tan ^2 \theta}}{1-\tan ^2 \theta}\right| \)
\( \Rightarrow \frac{\lambda^2}{4}+\tan ^2 \theta=\tan ^2 \theta \)
\( \Rightarrow \lambda=0\)
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