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Find the perpendicular distance from the origin to 6=r(\cos \theta +{\sqrt  {3}}\sin \theta )\,

Given line is 6=r(\cos \theta +{\sqrt  {3}}\sin \theta )\,

The line in the normal form is p=r\cos(\theta -\alpha )=r\cos \theta \cos \alpha +r\sin \theta \sin \alpha \,

The two equations represent the same line


{\frac  {p}{6}}={\frac  {\cos \alpha }{1}}={\frac  {\sin \alpha }{{\sqrt  {3}}}}\,

\cos \alpha ={\frac  {p}{6}},\sin \alpha ={\frac  {{\sqrt  {3}}p}{6}}\,

Squaring and adding both sides,we get

{\frac  {p^{2}}{36}}+{\frac  {3p^{2}}{36}}=1,{\frac  {4p^{2}}{36}}=1,p^{2}=9,p=3\,

Therefore the length of the perpendicular from the origin to the given line is 3.

Main Page:Geometry:Polar Coordinates