Question

Convert the rectangular equation to polar form and sketch its graph. $$ \left(x^{2}+y^{2}\right)^{2}-9\left(x^{2}-y^{2}\right)=0 $$

Short answer

The polar representation of the given rectangular equation is \( r = 0 \) and \( r = \pm 3\sqrt{\cos(2\theta)} \). The graph represents a lemniscate with a maximum distance of 3 from the origin along the x-axis.

Step-by-step solution

Convert to Polar Coordinates

Substitute \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\) into the given equation. This gives us \( \left(r\cos(\theta)^{2}+r\sin(\theta)^{2}\right)^{2}-9\left(r\cos(\theta)^{2}-r\sin(\theta)^{2}\right)=0 \)

Simplify using Trigonometric Identities

Knowing that \( \cos^2(\theta) + \sin^2 (\theta) = 1 \) and \( \cos^2 (\theta) - \sin^2 (\theta) = \cos (2\theta) \), the equation simplifies to \( r^4 - 9r^2\cos(2\theta) = 0 \)

Factor Out Common Term

Factor out the common term \(r^2\), which simplifies the equation to \( r^2(r^2 - 9\cos(2\theta)) = 0 \). The solutions are \( r = 0 \), and \( r^2 = 9\cos(2\theta) \) or \( r = \pm 3\sqrt{\cos(2\theta)} \)

Sketch the Graph

We know that \( r = 0 \) represents the origin. The other equation represents a lemniscate. As \( \cos(2\theta) \) varies between -1 and 1, the graph will extend from -3 to 3 in both the x and y-directions. The graph has two loops, one on either side of the origin, each with a maximum distance of 3 from the origin along the x-axis