How do you convert #x^2 - y^2 = 5# in polar form?

1 Answer
Feb 11, 2016

The polar form of that equation will be r= #sqrt[5/cos(2θ)]#.

Explanation:

To solve this problem, you have to understand the relationship that x and y has with r. r is the radius and has a starting point on the origin with an ending point anywhere on the graph. However, to use r in math, you have to break it into components since the angle provides challenges with calculations. The x component is how far away it is left or right from the origin. The y component is how far it is up or down from the origin. Together, the x and y components with r make a right triangle like so:

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The x component is the adjacent side to the angle, so use cosine (Cosine#theta# equals adjacent side divided by the hypotenuse or in this case, r.) to make the equation cos#theta# = x/r. To find x, multiple both sides by r to get
x= rcos#theta#

The y component is the opposite side to the angle, so use sine (Sine#theta# equals opposite side divided by the hypotenuse, r.) This yields the equation sin#theta# = y/r. Multiple both sides by r to simplify.
y = rsin#theta#

Now that you know what x and y equals, plug them in to the equation.

#(rcosθ)^2# - #(rsinθ)^2#= 5
Square both sides to get
#r^2cos^2θ# - #r^2sin^2θ#= 5
Factor out the #r^2# because it is a common factor
#r^2#(#cos^2θ# - #sin^2θ#)= 5
Using the trigonometry identity of cos(2θ)= #cos^2θ# - #sin^2θ#
(just have to memorize that identity)
#r^2#[cos(2θ)]= 5
Divide both sides by cos(2θ) to isolate r.
#r^2#= 5/(cos2θ)
Square root it
r= #sqrt[5/cos(2θ)]#

However since a square root can't be negative, there are some limits to the domain of cos(2θ).
Normally, cosθ is positive from -#pi#/2 < θ < #pi#/2.
However, since the angle is doubled in cos(2θ), divide by two.
This yields the final domain for this function to be -#pi#/4 < θ < #pi#/4