91Ó°ÊÓ

The given polar equations are \( r = 6 \sin \theta \) and \( r = \frac{6}{1 + 2 \sin \theta} \). Let's convert them into rectangular coordinates to make them easier to graph. For the first equation, we know: \[x = r \cos \theta, \, y = r \sin \theta.\] Substituting \( y = r \sin \theta = 6 \sin \theta \), we get \( y = 6 \sin \theta \). Further substitution with \( \sin \theta = \frac{y}{r} \), gives \( y = 6 \left(\frac{y}{r}\right) \) or \( r = 6 \). Rearrange to get \[ x^2 + y^2 = 6. \]Convert the second equation similarly: \( r = \frac{6}{1 + 2 \sin \theta} \), using \( r = \sqrt{x^2 + y^2} \), and \( \sin \theta = \frac{y}{\sqrt{x^2 + y^2}} \), we get: \[ r (1 + 2 \sin \theta) = 6, \]using \( r = \sqrt{x^2 + y^2} \), substitute into:\[ 1 + \frac{2y}{\sqrt{x^2 + y^2}} = \frac{6}{\sqrt{x^2 + y^2}}. \]

We obtained the following equations: 1. \( x^2 + y^2 = 36 \) from \( r = 6 \sin \theta \).2. \( \sqrt{x^2 + y^2} + 2y = 6 \) from \( r = \frac{6}{1 + 2 \sin \theta} \).Replace \( \sqrt{x^2 + y^2} \) in the second equation with \( r = \sqrt{x^2 + y^2} \), so:\[ \sqrt{x^2 + y^2} = 6 - 2y. \] Square both sides:\[ x^2 + y^2 = (6 - 2y)^2. \] Now expand:\[ x^2 + y^2 = 36 - 24y + 4y^2. \]Combine this with \( x^2 + y^2 = 36 \), and subtract:\[ 0 = -24y + 4y^2, \] giving us \( 4y^2 - 24y = 0 \). This can be factored as \[ 4y(y - 6) = 0 \].

Solving the factored equation \( 4y(y - 6) = 0 \) gives us \( y = 0 \) or \( y = 6 \). For \( y = 0 \), substitute back into either original equation:\( x^2 + 0^2 = 36 \) gives \( x = \pm 6 \), so points (6,0) and (-6,0).For \( y = 6 \), substitute:\( x^2 + 36 = 36 \) giving \( x = 0 \), so point (0,6).

Graph both equations in polar form to visually verify the solutions. The curve \( r = 6 \sin \theta \) is a circle, symmetric about y-axis with a center at \( (3, 3) \) in polar form, while \( r = \frac{6}{1 + 2 \sin \theta} \) represents a limaçon. Check the intersections by graphing them with the resulting intersection points: (6,0), (-6,0), and (0,6).