MHT CET · Maths · Pair of Lines
If \(\varphi^{\prime}\) is the angle between the lines \(a x^{2}+2 h x y+b y^{2}=0\), then angle between \(x^{2}+2 x y \sec \theta+y^{2}=0\) is
- A \(\bar{\theta}\)
- B \(2 \theta\)
- C \(\frac{\theta}{2}\)
- D \(3 \theta\)
Answer & Solution
Correct Answer
(A) \(\bar{\theta}\)
Step-by-step Solution
Detailed explanation
Angle between the lines \(a x^{2}+2 h x y+b y^{2}=0\) is \(\tan \theta=\left|\frac{2 \sqrt{h^{2}-a b}}{a+b}\right|\)
For \(x^{2}+2 x y \sec \theta+y^{2}=0\)
\(\begin{aligned} h &=\sec \theta, a=b=1 \\ \therefore \quad \tan \phi &=\mid \frac{2 \sqrt{\sec ^{2} \theta-1}}{1+1} \end{aligned}\)
\(=\frac{2 \tan \theta}{2}=\tan \theta\)
\(\therefore\) Angle between \(x^{2}+2 x y \sec \theta+y^{2}=0\) is \(\theta\).
For \(x^{2}+2 x y \sec \theta+y^{2}=0\)
\(\begin{aligned} h &=\sec \theta, a=b=1 \\ \therefore \quad \tan \phi &=\mid \frac{2 \sqrt{\sec ^{2} \theta-1}}{1+1} \end{aligned}\)
\(=\frac{2 \tan \theta}{2}=\tan \theta\)
\(\therefore\) Angle between \(x^{2}+2 x y \sec \theta+y^{2}=0\) is \(\theta\).
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