Problem 25 - 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 \(\frac{y^{2}}{16}-\frac{x^{2}}{4}=1\)

Step-by-Step Solution

Answer

The center of the hyperbola is at (0,0). The vertices are at (0,4) and (0,-4). The foci are at (0,\(\sqrt{20}\)) and (0,-\(\sqrt{20}\)). The equations of the asymptotes are y=±2x.

Step 1: Determine the center

From the equation \(\frac{y^{2}}{16}-\frac{x^{2}}{4}=1\), since there is no term deducted or added to either x or y, the center of the hyperbola is at the origin, i.e, (0,0).

Step 2: Find the vertices

The square root of the denominator under \(y^{2}\) gives us 'a', which is 4. Therefore, the vertices of the hyperbola are at (0, 0±4) = (0,4) and (0,-4).

Step 3: Determine the foci

First, find 'b' which is the square root of denominator under \(x^{2}\) = 2. Then, calculate the value of 'c' which is \(\sqrt{a^{2} + b^{2}} = \sqrt{4^{2} + 2^{2}} = \sqrt{20}\). Therefore, the foci are at (0,0±\sqrt{20}) = (0,\(\sqrt{20}\)) and (0,-\(\sqrt{20}\)).

Step 4: Find the asymptotes

The equations of asymptotes for the given hyperbola are y=± \(\frac{a}{b}\)x = ± \(\frac{4}{2}\)x = ±2x.

Step 5: Sketch the hyperbola

Start by placing the center, vertices and foci found above on the graph. Then, draw the asymptotes. Using these as a guide, sketch the branches of the hyperbola.

Key Concepts

Center of Hyperbola

In hyperbola equations like \(\frac{y^{2}}{16}-\frac{x^{2}}{4}=1\), determining the center is straightforward. Here, since there are no additional terms added or subtracted to the \(x\) or \(y\) variables, the center is at the origin, specifically at \((0,0)\). It serves as the symmetrical midpoint from which the hyperbola's features are balanced. To locate the center, look for any shifts in the \(x\) or \(y\) terms of the standard form, as those would indicate a different center. In this case, no shifts mean the center remains at the origin, making it a fundamental starting point for sketching the hyperbola.

Vertices of Hyperbola

The vertices of a hyperbola are crucial points that lie on the hyperbola along its transverse axis. For the equation \(\frac{y^{2}}{16}-\frac{x^{2}}{4}=1\), we find 'a' by taking the square root of the denominator under \(y^{2}\). Here, \(a = 4\). This value tells us how far the vertices are from the center along the \(y\)-axis. Thus, the vertices are at \((0, 4)\) and \((0, -4)\). These vertices help to define the shape and orientation of the hyperbola. They lie on the main axis of the hyperbola, confirming that the hyperbola opens vertically.

Foci of Hyperbola

Understanding the foci of a hyperbola is key to comprehending its reflective property. Foci are located further out along the transverse axis compared to the vertices. First, find the value of 'b' by taking the square root of the denominator under \(x^{2}\), which in this case is 2. Then, calculate 'c' using the formula \(c = \sqrt{a^{2} + b^{2}}\). So, \(c = \sqrt{16 + 4} = \sqrt{20}\). The foci are thus positioned at \((0, \sqrt{20})\) and \((0, -\sqrt{20})\). The foci are significant because distances from any point on the hyperbola to the foci maintain a constant difference, affirming the hyperbola's unique geometry.

Asymptotes of Hyperbola

Asymptotes are the invisible guidelines that dictate the hyperbola's open end directions. For this hyperbola, asymptotes can be calculated using the formula \(y = \pm \frac{a}{b}x\). Substituting \(a = 4\) and \(b = 2\), we get the asymptotes as \(y = \pm 2x\). These straight lines intersect at the hyperbola's center and guide its gradual curve pattern. Placing these lines on the graph before sketching helps in accurately portraying how both branches of the hyperbola approach the asymptotes but never meet them. Asymptotes act as boundaries indicating the direction and steepness of the hyperbola as it extends to infinity.