how do you convert 2x^2-2y^2=5y to polar form?

x = r * cos(t)

y = r * sin(t)

2 * (r * cos(t))^2 - 2 * (r * sin(t))^2 = 5 * r * sin(t)

2 * r^2 * cos(t)^2 - 2 * r^2 * sin(t)^2 = 5r * sin(t)

2r^2 * (cos(t)^2 - sin(t)^2) = 5r * sin(t)

2r^2 * cos(2t) = 5r * sin(t)

r^2 = 5 * r * sin(t) / (2 * cos(2t))

r = 5 * sin(t) / (2 * cos(2t))

r = (5/2) * (sin(t) / cos(2t))

well, x=rcost, and y=rsint in polar coordinates. (r=radius, t=theta)

i would just simply plug in those two into the given equation

2*(rcost)^2-2*(rsint)^2 = 5rsint

2*r^2*((cost)^2 - (sint)^2) = 5*r*sint

2*r*((cost)^2 - (sint)^2) = 5*sint

that's about the best i can do, hope this helps

x = r cos(t)

y = r sin(t)

2(r cos(t))^2 - 2(r sin(t))^2 = 5(r sin(t))

2 r^2 cos^2(t) - 2 r^2 sin^2(t) = 5r sin(t)

2r^2 (cos^2 (t) - sin^2 (t)) = 5r sin(t)

x = RcosΘ

y = RsinΘ

2x² - 2y² = 5y

2R²cos²Θ - 2R²sin²Θ = 5RsinΘ

2R²(cos²Θ - sin²Θ = 5RsinΘ

2R(1 - 2sin²Θ) = 5sinΘ

2R = 5sinΘ/(1 - 2sin²Θ)

R = 5sinΘ/(2cos2Θ)

...

are you doing summer work for mr. jones bc calc?