Problem 36 - College Algebra

Question

Use the center, vertices, and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. $$\frac{(x+2)^{2}}{9}-\frac{y^{2}}{25}=1$$

Step-by-Step Solution

Answer

The hyperbola is centered at (-2,0) with vertices at (-5,0) and (1,0). The asymptotes have equations \(y= ± \frac{5}{3}(x+2)\). The foci are at \((-2±\sqrt{34},0)\).

Step 1: Identify the center, a, and b

From the given equation, we can see that we have an hyperbola centered at (-2,0) (the opposites of the values being subtracted in the brackets are taken as the coordinates of the center). From the denominators of the fractions, it can be inferred that \(a^{2}=9\) meaning that \(a=3\) and \(b^{2}=25\) giving \(b=5\). These values correspond to distances from the center to the vertices and ends of the minor axis bi-sected by the center respectively.

Step 2: Plot the center, vertices and asymptotes

The center of the hyperbola is at (-2,0). The vertices of the hyperbola are \(a\) units to the left and right of the center at (-2±3,0) which corresponds to points (-5,0) and (1,0). The asymptotes of the hyperbola are given by the equation \(y=k ± \frac{b}{a}(x-h)\), where \((h,k)\) is the center. This corresponds to \(y=± \frac{5}{3}(x+2)\). The asymptotes intersect the x-axis at \((h±ab,0)\), corresponding to \(-2±3*5=-2±15=(-17,0) and (13,0)\). Advance up and down the asymptotes from the point of intersection, plot and sketch the asymptotes.

Step 3: Compute and plot the foci

The foci are found at a distance of \(c\) units to the right and left of the center, where \(c = \sqrt{a^{2}+b^{2}}\). Substituting results in \(c = \sqrt{9+25} = \sqrt{34}\). Therefore, the foci will be located at \((-2±\sqrt{34},0)\). Plot these points as well on the graph.