# X 2 Y 2 5

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

Source: https://socratic.org/questions/how-do-you-convert-x-2-y-2-5-in-polar-form

Posted by: soaltugas.net