Question

Graph the following conic sections, labeling the vertices, foci, directrices, and asymptotes (if they exist). Use a graphing utility to check your work. $$r=\frac{6}{3+2 \sin \theta}$$

Short answer

Answer: The polar equation \(r=\frac{6}{3+2\sin\theta}\) represents an ellipse. Its properties are as follows: - Vertices: \((\pm\sqrt{3}, 3)\) - Foci: \((\pm\sqrt{5/2}, 3)\) - Directrices: \(x = \pm\frac{3}{\sqrt{5/2}}\)

Step-by-step solution

Convert the polar equation to a Cartesian equation

In order to convert the given polar equation $$r=\frac{6}{3+2 \sin \theta}$$ into a Cartesian equation, we can use the following conversion formulas: $$ x = r\cos \theta \\ y = r\sin \theta $$ We can also use the Pythagorean relationship in polar coordinates, which is \(r^2 = x^2 + y^2\). Using these formulas, we can rewrite the polar equation in terms of x and y. $$ r = \frac{6}{3+2 \sin \theta} \\ \Rightarrow r(3+2\sin\theta) = 6 $$ Now, substitute \(x = r\cos\theta\) and \(y = r\sin\theta\) and simplify the equation. $$ r(3+2\frac{y}{r}) = 6 \\ \Rightarrow r^2(3+\frac{2y}{r}) = 6r \\ \Rightarrow (x^2+y^2)(3+\frac{2y}{\sqrt{x^2+y^2}}) = 6\sqrt{x^2+y^2} $$

Identify the type of conic section and its properties

Upon simplification and rearranging the terms, the Cartesian equation takes the form: $$ (2y^2-6y+x^2)+3x^2 = 0 $$ At this step, we realize that the equation represents an ellipse, with the given equation being the general form for ellipse: $$ \frac{x^2}{a^2} + \frac{(y-h)^2}{b^2} = 1 $$ Now, we find the values of \(a\), \(b\), and \(h\), and use them to find the ellipse's properties such as vertices, foci, directrices, and asymptotes.

Find the values of a, b, and h

Comparing the general form of an ellipse with the given equation, we get the values of \(a\), \(b\), and \(h\): $$ a^2 = 3 \\ b^2 = \frac{1}{2} \\ h = 3 $$ Now that we have the values of \(a\), \(b\), and \(h\), we can find the properties of the ellipse.

Find the vertices, foci, and directrices of the ellipse

Vertices: The vertices of the ellipse are the points \((\pm a, h)\), which are \((\pm\sqrt{3}, 3)\) in this case. Foci: The distance from the center of the ellipse to each focus is given by $$c = \sqrt{a^2 - b^2}$$; so, the foci can be located at \((\pm c, h)\), which are \((\pm\sqrt{5/2}, 3)\) in this case. Directrices: The directrices are vertical lines since it's a horizontal ellipse, given by the equation \(x = \pm\frac{a^2}{c}\), which are \(x = \pm\frac{3}{\sqrt{5/2}}\). Note: Asymptotes do not exist for an ellipse; they are only applicable to hyperbolas.

Graph the conic section and verify with a graphing calculator

After obtaining the properties of the ellipse, we can plot the points (vertices, foci, and directrices) on the graph and draw the ellipse. We can then verify our work using a graphing utility such as Desmos or Geogebra by entering the given equation in polar coordinates: $$ r=\frac{6}{3+2 \sin \theta} $$ The graph should align with our findings (vertices, foci, and directrices).