A) \[2=\sqrt{3}\,r\cos \theta -2\,r\sin \theta \]
B) \[5=-2\sqrt{3}\,r\sin \theta +4\,r\cos \theta \]
C) \[2=\sqrt{3}\,r\cos \theta +2\,r\cos \theta \]
D) \[5=2\sqrt{3}\,r\sin \theta +4\,r\cos \theta \]
Correct Answer: A
Solution :
Perpendicular to \[\sqrt{3}\sin \theta +2\cos \theta =\frac{4}{r}\] is \[\sqrt{3}\sin \left( \frac{\pi }{2}+\theta \right)+2\cos \left( \frac{\pi }{2}+\theta \right)=\frac{k}{r}\] It is passing through \[(-1,\,\pi \,/\,2)\] \[\sqrt{3}\sin \pi +2\cos \pi =\frac{k}{-1}\Rightarrow k=2\] \ \[\sqrt{3}\cos \theta -2\sin \theta =\frac{2}{r}\]Þ \[2=\sqrt{3}r\cos \theta -2r\sin \theta \].
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