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Question: The line passing through \(( - 1 , \pi / 2 )\) and perpendicular to \(\sqrt { 3 } \sin \theta + 2 \...

The line passing through (−1,π/2)( - 1 , \pi / 2 ) and perpendicular to 3sin⁡θ+2cos⁡θ=4r\sqrt { 3 } \sin \theta + 2 \cos \theta = \frac { 4 } { r } is.

A

2=3rcos⁡θ−2rsin⁡θ2 = \sqrt { 3 } r \cos \theta - 2 r \sin \theta

B

5=−23rsin⁡θ+4rcos⁡θ5 = - 2 \sqrt { 3 } r \sin \theta + 4 r \cos \theta

C

2=3rcos⁡θ+2rcos⁡θ2 = \sqrt { 3 } r \cos \theta + 2 r \cos \theta

D

5=23rsin⁡θ+4rcos⁡θ5 = 2 \sqrt { 3 } r \sin \theta + 4 r \cos \theta

Answer

2=3rcos⁡θ−2rsin⁡θ2 = \sqrt { 3 } r \cos \theta - 2 r \sin \theta

Explanation

Solution

Perpendicular to 3sin⁡θ+2cos⁡θ=4r\sqrt { 3 } \sin \theta + 2 \cos \theta = \frac { 4 } { r } is

3sin⁡(π2+θ)+2cos⁡(π2+θ)=kr\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,π/2)( - 1 , \pi / 2 )

3sin⁡π+2cos⁡π=k−1⇒k=2\sqrt { 3 } \sin \pi + 2 \cos \pi = \frac { k } { - 1 } \Rightarrow k = 2

∴ 3cos⁡θ−2sin⁡θ=2r\sqrt { 3 } \cos \theta - 2 \sin \theta = \frac { 2 } { r } ⇒ 2=3rcos⁡θ−2rsin⁡θ2 = \sqrt { 3 } r \cos \theta - 2 r \sin \theta.