MHT CET · Maths · Pair of Lines
The angle between the lines \(y^{2} \sin ^{2} \theta-x y \sin ^{2} \theta+x^{2}\left(\cos ^{2} \theta-1\right)=0\) is
- A \(\frac{\pi}{4}\)
- B \(\frac{\pi}{3}\)
- C \(\frac{\pi}{6}\)
- D \(\frac{\pi}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{\pi}{2}\)
Step-by-step Solution
Detailed explanation
We have \(y^{2} \sin ^{2} \theta-x y \sin ^{2} \theta+x^{2}\left(\cos ^{2} \theta-1\right)=0\)
\(\therefore\left(\sin ^{2} \theta\right) y^{2}-\left(\sin ^{2} \theta\right)(x y)-\left(\sin ^{2} \theta\right) x^{2}=0\)
\(\therefore\left(\sin ^{2} \theta\right)\left(y^{2}-x y-x^{2}\right)=0 \quad \Rightarrow \sin ^{2} \theta=0 \Rightarrow \theta=\frac{\pi}{2}\)
Note : sum of coefficients of \(x^{2}\) and \(y^{2}\) is zero.
\(\therefore\left(\sin ^{2} \theta\right) y^{2}-\left(\sin ^{2} \theta\right)(x y)-\left(\sin ^{2} \theta\right) x^{2}=0\)
\(\therefore\left(\sin ^{2} \theta\right)\left(y^{2}-x y-x^{2}\right)=0 \quad \Rightarrow \sin ^{2} \theta=0 \Rightarrow \theta=\frac{\pi}{2}\)
Note : sum of coefficients of \(x^{2}\) and \(y^{2}\) is zero.
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