Problem 34 - College Algebra

Question

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

Step-by-Step Solution

Answer

The center of the hyperbola is \((-2, 1)\), vertices are \((1, 1)\) and \((-5, 1)\), foci are \((-2+\sqrt{34}, 1)\) and \((-2-\sqrt{34}, 1)\), and the equations of the asymptotes are \(y = 1 \pm \frac{5}{3}(x+2)\).

Step 1: Identify the center

The center of the hyperbola is obtained from the equation as \((-h, -k)\) or \((-2, 1)\).

Step 2: Find the vertices

The vertices of the hyperbola are at a distance of a (the square root of the denominator under the x-term in the equation) units horizontally from the center because our hyperbola is oriented horizontally. Here, \(a = \sqrt{9} = 3\), so the vertices are \((-2+3, 1)\) and \((-2-3, 1)\) or \((1, 1)\) and \((-5, 1)\).

Step 3: Compute the foci

For any hyperbola, the distance from the center to each focus is given by \(c = \sqrt{a^2 + b^2}\). In our equation, \(a = 3\) and \(b = \sqrt{25} = 5\), so \(c= \sqrt{3^2+5^2} = \sqrt{34}\). The foci are therefore at a distance of \(\sqrt{34}\) units horizontally from the center because our hyperbola is oriented horizontally. So the foci are at \((-2+\sqrt{34}, 1)\) and \((-2-\sqrt{34}, 1)\).

Step 4: Determine the equations of the asymptotes

The equations of the asymptotes of any hyperbola in the general form are given by \(y = k \pm \frac{b}{a}(x-h)\). Plugging in \(h = -2\), \(k = 1\), \(a = 3\), and \(b = 5\), we have the asymptote equations \(y = 1 \pm \frac{5}{3}(x+2)\).