How do you find the polar equation for #2x^2+2y^2+5x=0#?

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Answer 1 · 2016-10-01T08:48:33

#r--2.5 cos theta# that represents the circle of radius 1.25, with center at #(1.25, pi)#.

The conversion equation is #(x, y) = r( cos theta, sin theta )#, giving

#r = sqrt ( x^2 + y^2 ), x = r cos theta and y = r sin theta#.

Here, .

#2(x^2+y^2)+5x=2r^2+5rcostheta =r(5r+2 cos theta)=0, giving, r = 0 and 5+2 cos theta =0#

The second includes r = o

at #theta = pi/2#, and again, at #theta = 3/2pi#

So, the polar form is r = -2,5 cos theta#

This represents the circle of radius 1.25, with center at #(1.25, pi)#.

For one complete circle, #theta in (pi/2. 3/2pi)#.