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, $theta in (pi/2. 3/2pi)$ .

How do you find the polar equation for 2x^2+2y^2+5x=02x^2+2y^2+5x=0?