What are the polar coordinates of #(x-1)^2-(y+5)^2=-24#?

1 Answer
Dec 17, 2016

Expand the squares, substitute #y = rsin(theta) and x = rcos(theta)#, and then solve for r.

Explanation:

Given: #(x - 1)^2 - (y + 5)^2 = -24#

Here is a graph of the above equation:

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Convert to polar coordinates.

Expand the squares:

#x^2 -2x + 1 - (y^2 + 10y + 25) = -24#

Regroup by power:

#x^2 - y^2 -2x - 10y + 1 - 25 = -24#

Combine the constant terms:

#x^2 - y^2 -2x - 10y = 0#

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

#(rcos(theta))^2 - (rsin(theta))^2 -2(rcos(theta)) - 10(rsin(theta)) = 0#

Lets move the factors of r outside the ():

#(cos^2(theta) - sin^2(theta))r^2 -(2cos(theta) + 10sin(theta))r = 0#

There are two roots, #r = 0# which is trivial should be discarded, and:

#(cos^2(theta) - sin^2(theta))r -(2cos(theta) + 10sin(theta)) = 0#

Solve for r:

#r = (2cos(theta) + 10sin(theta))/(cos^2(theta) - sin^2(theta))#

Here is the graph of the above equation:

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