Problem 30 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the center, vertices, foci, and asymptotes of the hyperbola, and sketch its graph using the asymptotes as an aid. Use graphing utility to verify your graph \(\frac{(y+4)^{2}}{\frac{1}{9}}-\frac{(x+3)^{2}}{\frac{1}{4}}=1\).

Step-by-Step Solution

Answer

The center of the hyperbola is at (-3,-4), the vertices are at (-3, -1) and (-3, -7), the foci are at (-3, -4 \(\pm\) \(\sqrt{13}\)), and the equations of the asymptotes are \(y = -4 + \frac{3}{2}(x + 3)\) and \(y = -4 - \frac{3}{2}(x + 3)\). Use the vertices, foci, and asymptotes to sketch the hyperbola on a graph.

Step 1: Identify Center of Hyperbola

The center of the hyperbola is given by the terms opposite the y and x in the equation. Therefore, the center will be at (-3, -4).

Step 2: Identify values of a, b and find c

In the equation, the terms under the y and x parts are reciprocals of \(a^2\) and \(b^2\) respectively. Therefore \(a^2 = 9\) and \(b^2=4\). Thus, \(a = 3\) and \(b = 2\). We can find the value of c (the distance from the center of the hyperbola to each focus) using the formula \(c = \sqrt{a^2 + b^2}\). Here \(c = \sqrt{9 + 4} = \sqrt{13}\).

Step 3: Determine the vertices and foci

The vertices of the hyperbola are \((h, k \pm a)\) or in this case (-3, -4 \(\pm\) 3). Therefore vertices are (-3, -1) and (-3, -7). The position of the foci is given by \((h, k \pm c)\). Hence, foci are located at (-3, -4 \(\pm\) \(\sqrt{13}\))

Step 4: Determine the asymptotes

The lines of the asymptotes are \(y = k \pm \frac{a}{b}(x-h)\). Substituting for the values of h, k, a, and b we find \(y = -4 \pm \frac{3}{2}(x+3)\). This gives two equations \(y = -4 + \frac{3}{2}(x + 3)\) and \(y = -4 - \frac{3}{2}(x + 3)\), which are the equations of the asymptotes.

Step 5: Sketch the Graph

Now plot the center, vertices, and foci on a graph. Sketch the asymptotes and draw the hyperbola approaching the asymptotes.

Key Concepts

Center of Hyperbola

Understanding where the center of a hyperbola lies is crucial to graphing it properly. The center is essentially the midpoint between the vertices and serves as a reference point for the entire graph. In the given equation, \(\frac{(y+4)^2}{\frac{1}{9}}-\frac{(x+3)^2}{\frac{1}{4}}=1\), we discover the center by looking at what’s added to \(x\) and \(y\) within the parentheses. Opposite these terms, the center of our hyperbola is at (-3, -4).

Vertices and Foci of Hyperbola

The vertices and foci of a hyperbola are points that define its shape and orientation. For our hyperbola, the vertices are located \(a\) units above and below the center along the vertical transverse axis, because \(a^2\) is under the \(y\)-term. With \(a = 3\), the vertices are at (-3, -4 \(\pm\) 3), resulting in the points (-3, -1) and (-3, -7).

Now, for the foci, which lie along the same axis but farther from the center: we calculate \(c = \sqrt{a^2 + b^2}\), resulting in \(c = \sqrt{13}\). The foci are then (-3, -4 \(\pm\) \(\sqrt{13}\)), which reveals just how much the hyperbola stretches outward from the center compared to its vertices.

Asymptotes of Hyperbola

Asymptotes are imaginary lines that the hyperbola approaches but never touches. They provide a skeleton that helps us sketch the curve accurately. To find the equations of the asymptotes for our hyperbola, we use the formula \(y = k \pm \frac{a}{b}(x-h)\). Plugging in the center coordinates \(h, k\) = (-3, -4) and the values \(a = 3\) and \(b = 2\), we derive the equations \(y = -4 \pm \frac{3}{2}(x + 3)\).

The two resulting lines, \(y = -4 + \frac{3}{2}(x + 3)\) and \(y = -4 - \frac{3}{2}(x + 3)\), guide us when drawing the hyperbola, as the branches must approach these lines as they extend towards infinity. Recognizing these asymptotes is essential for visualizing the hyperbola's open, divergent nature.