TS EAMCET · Maths · Pair of Lines
The difference of the tangents of the angles which the lines \(\left(\tan ^2 \alpha+\cos ^2 \alpha\right) x^2-2 x y \tan \alpha\) \(+\left(\sin ^2 \alpha\right) y^2=0\) make with the \(X\)-axis is
- A \(\frac{1}{2}\)
- B \(1\)
- C \(2\)
- D \(\frac{\sqrt{3}-1}{2}\)
Answer & Solution
Correct Answer
(C) \(2\)
Step-by-step Solution
Detailed explanation
We have, \(x^2\left(\tan ^2 \alpha+\cos ^2 \alpha\right)-2 x y \tan \alpha+y^2 \sin ^2 \alpha=0\) Comparing this equation with \(a x^2+2 h x y+b y^2=0\), We have, \(a=\tan ^2 \alpha+\cos ^2 \alpha, 2 h=-2 \tan \alpha\) and \(b=\sin ^2 \alpha\) Let \(m_1\) and \(m_2\) be the…
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