AP EAMCET · Maths · Pair of Lines
The angle between the lines \(a b\left(x^2-y^2\right)+\left(a^2-b^2\right) x y=0\) is
- A \(\frac{\pi}{2}\)
- B \(\frac{\pi}{3}\)
- C \(\frac{\pi}{4}\)
- D \(\frac{\pi}{6}\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi}{2}\)
Step-by-step Solution
Detailed explanation
We have, \[ \begin{aligned} & a b\left(x^2-y^2\right)+\left(a^2-b^2\right) x y=0 \\ & a b x^2+\left(a^2-b^2\right) x y-a b y^2=0 \end{aligned} \] Sum of the coefficient of \(x^2\) and \(y^2\) \[ \text { i.e. } a b-a b=0 \] \(\therefore \quad\) Lines are perpendicular to each…
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