The problem asks to convert the polar equation $r = \frac{4}{1 + 2\sin\theta}$ to a rectangular equation.

1. Problem Description

The problem asks to convert the polar equation $r = \frac{4}{1 + 2\sin\theta}$ to a rectangular equation.

2. Solution Steps

First, multiply both sides of the equation by $1 + 2\sin\theta$:

$r(1 + 2\sin\theta) = 4$

$r + 2r\sin\theta = 4$

Now, we use the following conversions:

$r = \sqrt{x^2 + y^2}$

$y = r\sin\theta$

Substitute these into the equation:

$\sqrt{x^2 + y^2} + 2y = 4$

Isolate the square root term:

$\sqrt{x^2 + y^2} = 4 - 2y$

Square both sides of the equation:

$(\sqrt{x^2 + y^2})^2 = (4 - 2y)^2$

$x^2 + y^2 = 16 - 16y + 4y^2$

Rearrange the terms to get a standard form:

$x^2 + y^2 - 4y^2 + 16y - 16 = 0$

$x^2 - 3y^2 + 16y - 16 = 0$

Complete the square for the $y$ terms:

$x^2 - 3(y^2 - \frac{16}{3}y) - 16 = 0$

$x^2 - 3(y^2 - \frac{16}{3}y + (\frac{8}{3})^2) - 16 + 3(\frac{8}{3})^2 = 0$

$x^2 - 3(y - \frac{8}{3})^2 - 16 + 3(\frac{64}{9}) = 0$

$x^2 - 3(y - \frac{8}{3})^2 - 16 + \frac{64}{3} = 0$

$x^2 - 3(y - \frac{8}{3})^2 + \frac{64 - 48}{3} = 0$

$x^2 - 3(y - \frac{8}{3})^2 + \frac{16}{3} = 0$

$x^2 - 3(y - \frac{8}{3})^2 = -\frac{16}{3}$

$3(y - \frac{8}{3})^2 - x^2 = \frac{16}{3}$

$\frac{(y - \frac{8}{3})^2}{\frac{16}{9}} - \frac{x^2}{\frac{16}{3}} = 1$

3. Final Answer

The rectangular equation is $x^2 - 3y^2 + 16y - 16 = 0$. Equivalently, the standard form for a hyperbola is $\frac{(y - \frac{8}{3})^2}{\frac{16}{9}} - \frac{x^2}{\frac{16}{3}} = 1$.