91Ó°ÊÓ

From the given equation \(r=\frac{5}{-1+2 \cos \theta}\), we start by identifying the type of conic. It can be inferred that this equation has the form \(r = \frac{p}{1 ± e cos(\theta - \varphi)}\) which is recognized as the polar equation for a conic section with eccentricity e. Here, if e < 1, the conic is an ellipse; if e = 1, it's a parabola; and if e > 1, it's a hyperbola. Here, the eccentricity e equals 2, indicating that the conic is a hyperbola.

For a hyperbola with a horizontal transverse axis, the standard form in Cartesian representation is \(\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\), where [h, k] are the coordinates of the center; a is the distance from the center to a vertex, b is the distance from the center to a co-vertex, and the vertices are located a units to the left and right of the center. In order to get the equation in a standard form, we also need to transform the equation to Cartesian coordinates. We use the formulas \(x = r cos( \theta)\) and \(y = r sin( \theta)\). So, we substitute \(r = x/cos( \theta)\) into the r given in the exercise. We then collect the terms and adjust the format of the equation. Let's denote a=√(-p/e) and b=√(-p/(e^2 - 1)) and we can find the center coordinates (h, k).

After getting the standard form, the center of the hyperbola, the values of a and b, we can sketch the graph. The center (h, k) is marked on the graph. Then, from the center, move a units horizontally (left or right depending on the equation) to plot the vertices, and b units vertically (up or down) to plot the co-vertices. Next, sketch a rectangle using these vertices and co-vertices as guide points. Then draw the asymptotes of the hyperbola, which are the diagonals of this rectangle. Finally, sketch the hyperbola branches approaching the asymptotes.