Question #1aa2d

Question

1956 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-27T02:24:02

$x^2+y^2=((x^2-y^2)/(x^2+y^2))^2$

Here is the graph of the original equation $r = cos(2theta)$ :

![Desmos.com]( useruploads.socratic.org )

Use the identity $cos(2theta) = cos^2(theta)-sin^2(theta)$

$r = cos^2(theta)-sin^2(theta)$ :

Substitute the $sqrt(x^2+y^2)$ for r, $x^2/(x^2+y^2)$ for $cos^2(theta)$ , and $y^2/(x^2+y^2)$ for $sin^2(theta)$ :

$sqrt(x^2+y^2)=x^2/(x^2+y^2)-y^2/(x^2+y^2)$

$sqrt(x^2+y^2)=(x^2-y^2)/(x^2+y^2)$

Here is the graph of that equation:

graph{sqrt(x^2+y^2)=x^2/(x^2+y^2)-y^2/(x^2+y^2) [-10, 10, -5, 5]}

It is only half of the loops

Square both sides:

$x^2+y^2=((x^2-y^2)/(x^2+y^2))^2$

Here is the graph of that equation:

graph{x^2+y^2=((x^2-y^2)/(x^2+y^2))^2 [-10, 10, -5, 5]}

This looks like the polar equation.