TS EAMCET · Maths · Three Dimensional Geometry
The polar equation of the line perpendicular to the line \(\sin \theta-\cos \theta=\frac{1}{r}\) and passing through the point \(\left(2, \frac{\pi}{6}\right)\) is
- A \(\sin \theta+\cos \theta=\frac{\sqrt{3}+1}{r}\)
- B \(\sin \theta-\cos \theta=\frac{\sqrt{3}+1}{r}\)
- C \(\sin \theta+\cos \theta=\frac{\sqrt{3}-1}{r}\)
- D \(\cos \theta-\sin \theta=\frac{\sqrt{3}}{r}\)
Answer & Solution
Correct Answer
(A) \(\sin \theta+\cos \theta=\frac{\sqrt{3}+1}{r}\)
Step-by-step Solution
Detailed explanation
Given polar equation of line \(\sin \theta-\cos \theta=1 / r\) ...(i) and point \((2, \pi / 6)\) Let \(\quad x=r \cos \theta=2 \cdot \cos \pi / 6=\sqrt{3}\) and \(y=r \sin \theta=2 \cdot \sin \pi / 6=1\) The cartesian point is \((\sqrt{3}, 1)\). Now, we change the polar of line…
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