Changing an equation from Rectangular to Polar

Question

So here's the question:

"Write the equation x² + xy + y² -2x +4y = -3 in polar coordinates and solve for r"

So so far I changed all the x's and y's to polar and used any identities I could to get the equation r² + r²cosθsinθ - 2rcosθ + 4rsinθ = -3 After that, I don't know how to solve for r nor do I know if this equation is really even right. Any help would be appreciated. Thanks in advance.

Answer 1

Hi Louis,

After switching to the polar coordinates through

x = r*cos(θ), y = r*sin(θ),

we obtain

(cos(theta)*sin(theta)+sin(theta)^2+cos(theta)^2) * r^2+2*(2*sin(theta)-cos(theta)) * r = -3.

Using the trigonometric identity

sin(theta)^2+cos(theta)^2 = 1,

our equation simplifies to

(cos(theta)*sin(theta)+1)*r^2+2*(2*sin(theta)-cos(theta))*r+3 = 0.

This is a quadratic equation for r, and is solved by using the following formulae.

Denoting the coefficient of r^2 by a, coefficient of r by b, and the free term by c, we first find the discriminant,

D = b^2 - 4*a*c = 4*(2*sin(theta)-cos(theta))^2 - 4*(cos(theta)*sin(theta)+1)*3 =

= 12*sin(theta)^2-28*cos(theta)*sin(theta)-8.

The two solutions for r are:

r = (-b + s*sqrt(D))/(2*a),

where

s=-1 or s=1.

This yields

r = 1/2*(-4*sin(theta)+2*cos(theta)+2*s*(3*sin(theta)^2-7*cos(theta)*sin(theta)-2)^(1/2))/(cos(theta)*sin(theta)+1).