MHT CET · Maths · Pair of Lines
If the pair of lines given by \((x \cos \alpha+y \sin \alpha)^2=\left(x^2+y^2\right) \sin ^2 \alpha\) are perpendicular to each other, then \(\alpha\) is
- A \(0\)
- B \(\frac{\pi}{2}\)
- C \(\frac{\pi}{4}\)
- D \(\frac{\pi}{6}\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
\((x \cos \alpha+y \sin \alpha)^2=\left(x^2+y^2\right) \sin ^2 \alpha \)
\( \therefore x^2 \cos ^2 \alpha+y^2 \sin ^2 \alpha+2 x y \sin \alpha \cos \alpha \)
\( =x^2 \sin ^2 \alpha+y^2 \sin ^2 \alpha \)
\( \therefore x^2\left(\cos ^2 \alpha-\sin ^2 \alpha\right)+2 x y \sin \alpha \cos \alpha=0 \)
This represents a pair of straight lines where \(\mathrm{a}=\cos ^2 \alpha-\sin ^2 \alpha, \mathrm{h}=\sin \alpha \cos \alpha\) and \(\mathrm{b}=0\)
As lines are perpendicular, we get
\(
\begin{aligned}
& a+b=0 \\
& \therefore \cos ^2 \alpha=\sin ^2 \alpha \Rightarrow \alpha=\frac{\pi}{4}
\end{aligned}
\)
\( \therefore x^2 \cos ^2 \alpha+y^2 \sin ^2 \alpha+2 x y \sin \alpha \cos \alpha \)
\( =x^2 \sin ^2 \alpha+y^2 \sin ^2 \alpha \)
\( \therefore x^2\left(\cos ^2 \alpha-\sin ^2 \alpha\right)+2 x y \sin \alpha \cos \alpha=0 \)
This represents a pair of straight lines where \(\mathrm{a}=\cos ^2 \alpha-\sin ^2 \alpha, \mathrm{h}=\sin \alpha \cos \alpha\) and \(\mathrm{b}=0\)
As lines are perpendicular, we get
\(
\begin{aligned}
& a+b=0 \\
& \therefore \cos ^2 \alpha=\sin ^2 \alpha \Rightarrow \alpha=\frac{\pi}{4}
\end{aligned}
\)
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