Problem 40 - College Algebra

Question

Use the center, vertices, and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. $$(x+3)^{2}-9(y-4)^{2}=9$$

Step-by-Step Solution

Answer

The hyperbola has a center at \(-3,4\), vertices at \((0,4)\) and \((-6,4)\), foci at \((-3+3\sqrt{2},4)\) and \((-3-3\sqrt{2},4)\), and the equations of the asymptotes are \(y = x + 7\) and \(y = -x + 1\).

Step 1: Identify the center, a and b

From the equation, we can see that the center of the hyperbola is at \(-3,4\). The value of \(a^2\) comes from the coefficient of \(x^2\), and the value of \(b^2\) comes from the coefficient of \(y^2\), so \(a^2 = 9\) and \(b^2 = -9\). This gives us \(a = 3\) and \(b = 3\).

Step 2: Identify and plot the vertices and center

The vertices of the hyperbola are at a distance \(a\) along the x-axis from the center. So they are at \((-3+a,4)\) and \((-3-a,4)\), which simplifies to \((0,4)\) and \((-6,4)\). Plot these points, along with the center, on the graph.

Step 3: Identify and plot the asymptotes

The equations of the asymptotes for a hyperbola centered at \((h,k)\) with a horizontal transverse axis are \(y = k \pm \frac{b}{a}(x-h)\). Given the center at \((-3,4)\), \(a = 3\), and \(b = 3\), this gives us the equations \(y = 4 \pm (x+3)\). In other words, \(y = x + 7\) and \(y = -x + 1\). These are the lines passing through the center that the hyperbola approaches as \(x\) goes to infinity or minus infinity. Plot these lines on the graph.

Step 4: Identify and plot the foci

The foci are at a distance of \(\sqrt{a^2 + b^2}\) from the center. This gives us the points \((-3+\sqrt{a^2 + b^2},4)\) and \((-3-\sqrt{a^2 + b^2},4)\), or \((-3+\sqrt{18},4)\) and \((-3-\sqrt{18},4)\), which simplify to \((-3+3\sqrt{2},4)\) and \((-3-3\sqrt{2},4)\). Plot these points on the graph.

Step 5: Sketch the hyperbola

With the center, vertices, foci, and asymptotes plotted, we can now sketch the hyperbola. It should be apparent that it opens to the left and right as it follows this formula \((x-h)^{2}/a^{2} - (y-k)^{2}/b^{2} = 1\), where \(a > b\). This implies the transverse axis is horizontal.