How do you convert #6=(5x-7y)^2-y+x# into polar form?

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Answer 1 · 2018-04-15T03:58:28

#color(indigo)(2r^2 sin theta (12 sin theta - 35 cos theta ) + r (cos theta - sin theta) + 19 = 0#

#x = r cos theta, y = r sin theta#

#(5x - 7y)^2 - y + x = 6#

#"Substituting values of x & y in terms of " r and theta#,

#(5 r cos theta - 7 r sin theta)^2 - r sin theta + cos theta = 6#

#25r^2 cos^2 theta + 49 r^2 sin^2 theta - 70 r^2 cos theta sin theta - r sin theta + r cos theta = 6#

#25 r^ cos^2 theta + 25 r^2 sin^2 theta + 24 r^ sin^2 theta - 70 r^2 cos theta sin theta - r sin theta + r cos theta = 6#

#25 + 24 r^2 sin^2 theta - 70 r^2 cos theta sin theta - r sin theta + r cos theta = 6#

#2r^2 sin theta (12 sin theta - 35 cos theta ) + r (cos theta - sin theta) + 19 = 0#