How do you convert #3=(x+4y)^2+3x# into polar form?

1 Answer
Nov 2, 2016

Please see the explanation for a detailed description of how the conversion is done.

Explanation:

Expand the square, using the F.O.I.L. method:

#3 = x^2 + 8xy + 16y^2 + 3x#

Substitute #rcos(theta) # for x and #rsin(theta)# for y:

#3 = (rcos(theta))^2 + 8(rcos(theta))(rsin(theta)) + 16(rsin(theta))^2 + 3(rcos(theta))#

This can be written as a quadratic in r:

#0 = (cos^2(theta) + 8cos(theta)sin(theta) + 16sin^2(theta))r^2 + 3cos(theta)r - 3#

The coefficient for #r^2# factors into a square:

#0 = (cos(theta) + 4sin(theta))^2r^2 + 3cos(theta)r - 3#

Use the quadratic formula:

#r = (-b +-sqrt(b^2 - 4(a)(c)))/(2a)#

where:

#a = (cos(theta) + 4sin(theta))^2#
#b = 3cos(theta)#
#c = -3#

Also, we must change the #+-# to only #+#, because negative values for r do not make sense.

#r = (-3cos(theta) +sqrt(9cos^2(theta) + 12(cos(theta) + 4sin(theta))^2))/(2(cos(theta) + 4sin(theta))^2)#