Question

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

Answer 1

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:

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