We will first consider the given polar coordinates, \(r =\frac{6}{2 + sin \theta}
(a) The objective is to write the equation in standard form.
Since we know that the standard form is as follows:
\(r = \frac{ep}{1 + e sin \theta}\)
Now, we will divide the numerator and denominator by 2.
\(r = \frac{\frac{6}{2}}{\frac{2}{2} + \frac{1}{2} sin \theta}\)

\(= 3/(1 + 1/2 sin \theta)\)

Hence, the required standard form is\)r = \frac{3}{1 + \frac{1}{2} sin \theta}\) (b) The next objective is to determine the values of e and p. On comparing with standard form, we get, \(e = \frac{1}{2}\) \(ep = 3 \Rightarrow p = 6\) Thus, the values are \(e= \frac{1}{6} and p = 6\). (c) Next, identify the conic section using the value of eccentricity. Since we know that the eccentricity of an ellipse which is not a circle is greater than zero but less than 1. Here, \(e = \frac{1}{2}\). Hence, we can conclude that the given conic equation is of ellipse.

\(= 3/(1 + 1/2 sin \theta)\)

Hence, the required standard form is\)r = \frac{3}{1 + \frac{1}{2} sin \theta}\) (b) The next objective is to determine the values of e and p. On comparing with standard form, we get, \(e = \frac{1}{2}\) \(ep = 3 \Rightarrow p = 6\) Thus, the values are \(e= \frac{1}{6} and p = 6\). (c) Next, identify the conic section using the value of eccentricity. Since we know that the eccentricity of an ellipse which is not a circle is greater than zero but less than 1. Here, \(e = \frac{1}{2}\). Hence, we can conclude that the given conic equation is of ellipse.