Home > Trigonometry > Converting Between Systems

How do you convert #y^2= 4(x+1) # in polar form?

Answer 1

Quinn Anderson

By converting cartesian to polar coordinates

Pose:

#x=rho cos(theta)# #y=rho sin(theta)#

The equation becomes:

#rho^2 sin^2(theta) = 4(rho cos(theta) + 1)#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Anna Andrews

To convert the equation (y^2 = 4(x + 1)) into polar form, you can use the following relationships: [x = r \cos(\theta)] [y = r \sin(\theta)] [r^2 = x^2 + y^2]

Substitute (x = r \cos(\theta)) and (y = r \sin(\theta)) into the equation (y^2 = 4(x + 1)), then solve for (r):

[(r \sin(\theta))^2 = 4(r \cos(\theta) + 1)] [r^2 \sin^2(\theta) = 4r \cos(\theta) + 4] [r^2 (\sin^2(\theta) - 4 \cos(\theta)) = 4] [r^2 = \frac{4}{\sin^2(\theta) - 4 \cos(\theta)}]

Therefore, the polar form of the equation is (r^2 = \frac{4}{\sin^2(\theta) - 4 \cos(\theta)}).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.