How do you graph #r=4/(-3+3sintheta)#?

1 Answer
Oct 15, 2016

This is a hyperbola with its center #(0, -4)#; its vertices are at #(0, -4 - sqrt(12)) and (0, -4 + sqrt(12))# and it opens downward and upward from its vertices.

Explanation:

Multiply both sides by the denominator:

#4rsin(theta) - 3r = 4#

Multiply both side by -1:

#3r - 4rsin(theta) = -4#

Move the sine term to the right:

#3r = 4rsin(theta)-4#

Factor out a 4:

#3r = 4(rsin(theta)-1)#

Square both sides:

#9r^2 = 16(rsin(theta)-1)^2#

Substitute #x^2 + y^2# for #r^2# and #y# for #rsin(theta)#

#9(x^2 + y^2) = 16(y - 1)^2#

Expand the square on the right:

#9(x^2 + y^2) = 16(y^2 - 2y + 1)#

distribute through the ()s:

#9x^2 + 9y^2 = 16y^2 - 32y + 16#

Combine like terms and leave the constant on the right:

#9x^2 - 4y^2 - 32y = 16#

Add #-4k^2# to both sides:

#9x^2 - 4y^2 - 32y -4k^2 = 16 - 4k^2#

Factor out -4 from the y terms

#9x^2 - 4(y^2 + 8y + k^2) = 16 - 4k^2#

Use 8y to find the value of #k# and #k^2#:

#-2ky = 8y#

#k = -4, k^2 = 16#

This makes the left a perfect square with k:

#9x^2 - 4(y - -4)^2 = -48#

Divide both sides by -48:

#(y - -4)^2/12 - 3/16x^2 = 1#

Put in standard form:

#(y - -4)^2/(sqrt(12))^2 - (x - 0)^2/(4sqrt(3)/3)^2 = 1#

This is a hyperbola with its center #(0, -4)# its vertices are at #(0, -4 - sqrt(12)) and (0, -4 + sqrt(12))#. It opens downward and upward from its vertices.