A) \( \Large 2=\sqrt{3}r\cos \theta-2r\sin \theta \) |
B) \( \Large 5=2\sqrt{3}r\sin \theta +4r\cos \theta \) |
C) \( \Large 2=\sqrt{3}r\cos \theta + 2r\sin \theta \) |
D) \( \Large 5=2\sqrt{3}r\sin \theta + 4r\cos \theta \) |
A) \( \Large 2=\sqrt{3}r\cos \theta-2r\sin \theta \) |
Any line which is perpendicular to
\( \Large \sqrt{3}\sin \theta + 2\cos \theta = \frac{4}{r} \) is
\( \Large \sqrt{3}\sin \left(\frac{ \pi }{2}+\theta \right) + 2\cos \left(\frac{ \pi }{2}+\theta \right)
= \frac{k}{r} \) ...(i)
Since, it is passing through \( \Large \left(-1,\ \frac{ \pi }{2}\right) \)
\( \Large \sqrt{3}\sin \pi + 2\cos \pi = \frac{k}{-1}=k=2 \)
Putting k = 2 in eq. (i) we get
\( \Large \sqrt{3}\cos \theta - 2\sin \theta = \frac{2}{r} \)
=> \( \Large 2 = \sqrt{3}r \cos \theta - 2r \sin \theta \)