AP EAMCET · Maths · Pair of Lines
If the slopes of both the lines given by \(x^2+2 h x y+6 y^2=0\) are positive and the angle between these lines is \(\operatorname{Tan}^{-1}\left(\frac{1}{7}\right)\), then the product of the perpendiculars
- A \(\frac{1}{6}\)
- B \(\frac{1}{5 \sqrt{2}}\)
- C \(\frac{5}{6}\)
- D \(\frac{1}{3 \sqrt{2}}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{5 \sqrt{2}}\)
Step-by-step Solution
Detailed explanation
\(\tan\theta = \frac{2\sqrt{h^2-ab}}{|a+b|} \implies \frac{1}{7} = \frac{2\sqrt{h^2-1 \cdot 6}}{|1+6|}\) \(1 = 2\sqrt{h^2-6} \implies \frac{1}{4} = h^2-6 \implies h^2 = \frac{25}{4}\) \(m_1+m_2 = -\frac{2h}{6} = -\frac{h}{3}\). Since…
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