91Ó°ÊÓ

To convert the polar equation \(r = \frac{1}{1 - \sin{\theta}}\) to a Cartesian equation, we will use the conversion formulas \(x = r \cos{\theta}\) and \(y = r \sin{\theta}\). First, let's solve for \(r\) in the equation \(y = r \sin{\theta}\). Since \(r = \frac{1}{1 - \sin{\theta}}\), we can substitute this in the equation for y: \(y = \frac{\sin{\theta}}{1 - \sin{\theta}}\). Now, we should find a relationship between x and y. Next, let's solve for \(r\) in the equation \(x = r \cos{\theta}\). Again, we substitute the polar equation for r: \(x = \frac{\cos{\theta}}{1 - \sin{\theta}}\). Now, we need to find a Cartesian equation relating x and y.

Divide the equations for y and x obtained in Step 1: \(\frac{y}{x} = \frac{\frac{\sin{\theta}}{1 - \sin{\theta}}}{\frac{\cos{\theta}}{1 - \sin{\theta}}}\) This simplifies to: \(\frac{y}{x} = \frac{\sin{\theta}}{\cos{\theta}}\) As we know, \(\frac{\sin{\theta}}{\cos{\theta}} = \tan{\theta}\), so we have: \(\frac{y}{x} = \tan{\theta}\) Now, remember from Step 1 that \(y = \frac{\sin{\theta}}{1 - \sin{\theta}}\). We can use the identity \(1 - \sin^2{\theta} = \cos^2{\theta}\), to find: \(1 - \sin{\theta} = \pm \sqrt{1 - \sin^2{\theta}} = \pm \cos{\theta}\) So, \(\tan{\theta} = \frac{y}{x} = \frac{\sin{\theta}}{\pm \cos{\theta}} = \frac{y}{1 - \sin{\theta}}\) Now we need to solve for \(\sin{\theta}\).

Solving the equation \(y = \frac{\sin{\theta}}{1 - \sin{\theta}}\) for \(\sin{\theta}\): \(\sin{\theta} = y(1 - \sin{\theta})\) \(y = y\sin{\theta}\) \(y - y\sin{\theta} = y\sin{\theta}\) So, \(\sin{\theta} = \frac{y}{2}\). Now, replacing \(\sin{\theta}\) in our Cartesian equation from Step 2: \(\tan{\theta} = \frac{y}{x} = \frac{\frac{y}{2}}{1 - \frac{y}{2}}\) Solve this equation for y: \(2x = y\left(2 - y\right)\) \(2x = 2y - y^2\) Rearrange to get the final Cartesian equation: \(y^2 - 2y + 2x = 0\)

Now, we have the Cartesian equation of the curve: \(y^2 - 2y + 2x = 0\) Move the terms associated with y to the left-hand side: \(y^2 - 2y + 1 = -2x + 1\) \((y - 1)^2 = -2(x - 1)\) The general form of a parabola with vertex at \((h, k)\) is: \((y - k)^2 = 4p(x - h)\) Comparing this to our equation \((y - 1)^2 = -2(x - 1)\), we find that the vertex is at \((h, k) = (1, 1)\) in polar coordinates. Now, let's convert this vertex back into Cartesian coordinates using \(x = r\cos{\theta}\) and \(y = r\sin{\theta}\): \(x = (1)(\cos(1)) = 0\), and \(y = (1)(\sin(1)) = -\frac{1}{2}\) So, the vertex of the parabola in Cartesian coordinates is \(\left(0, -\frac{1}{2}\right)\). The plot of the given polar equation is indeed a parabola with the vertex at \(\left(0, -\frac{1}{2}\right)\).