AP EAMCET · Maths · Straight Lines
If the coordinate axes are rotated through an angle \(\frac{\pi}{6}\) about the origin, then the transformed equation of \(\sqrt{3} x^2-4 x y+\sqrt{3} y^2=0\) is
- A \(\sqrt{3} y^2+x y=0\)
- B \(x^2 - y^2 = 0\)
- C \(\sqrt{3} y^2-x y=0\)
- D \(\sqrt{3} y^2- 2x y=0\)
Answer & Solution
Correct Answer
(C) \(\sqrt{3} y^2-x y=0\)
Step-by-step Solution
Detailed explanation
Given equation, \(\sqrt{3} x^2-4 x y+\sqrt{3} y^2=0\) Coordinate axes are rotated through an angle of \(\frac{\pi}{6}\) about the origin Now, axis is \(x \cos \frac{\pi}{6} \quad y \cos \frac{\pi}{6}\) and \(x \cos \frac{\pi}{6}+y \cos \frac{\pi}{6}\) Thus, equation becomes…
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