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

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: 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