Convert to polar form, solve for r y = 4x^2

Question

Hi, Can someone help me with this question?

I got an answer to = sin(theta)/ 6 x (cos(theta))^2

But my computer is saying this is wrong!

Am I am doing this wrong? Please provide an answer that works!

Answer 1

y=4x^2

y=rsin(theta), x=rcos(theta)

rsin(theta = 4(rcos(theta))^2 = 4r^2(cos^2(theta))

divide by r to get

sin(theta) = 4rcos^2(theta)

4r = sin(theta)/cos^2(theta)

r = sin(theta)/4cos^2(theta) or (1/4)tan(theta)sec(theta)

You may have it right except you have a 6 where you should have a 4

UNLESS the problem was really y=(4x)^2 = 16x^2

then r = sin(theta)/16cos^2(theta) and you had it right except for having a 6 where it should have been 16