Problem 29 - 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 \(\frac{(y+6)^{2}}{1}-\frac{(x-2)^{2}}{\frac{1}{16}}=1\)

Step-by-Step Solution

Answer

The hyperbola has center at (2, -6), vertices at (2, -5) and (2, -7), foci at (2, -6 ± \( \sqrt{17}/4 \)), and asymptotes with the equations y = 4x - 14 and y = -4x + 2.

Step 1: Find the center

The center of the hyperbola is given by the coordinates (h, k). Looking at the given equation \(\frac{(y+6)^{2}}{1}-\frac{(x-2)^{2}}{\frac{1}{16}}=1\), we can see that h = 2 and k = -6. So the center is (2, -6).

Step 2: Find the vertices

The vertices are located a units above and below the center. Here a^2 = 1 from the denominator of the y term. Therefore, a = 1. The vertices are at (2, -6 ±1) or (2, -5) and (2, -7).

Step 3: Find the foci

The foci are located c units above and below the center, where c^2 = a^2 + b^2. Here, b^2 = 1/16 from the denominator of the x term. So, b = 1/4. Therefore, c^2 = 1 + 1/16 = 17/16 and c = \( \sqrt{17}/4 \). The foci are at (2, -6 ± \( \sqrt{17}/4 \)).

Step 4: Find the asymptotes

The asymptotes are the lines passing through the center with slopes ±a/b. Slope is given by ±a/b = ±4. Therefore, the asymptote equations are y = 4x + k - 4h and y = -4x + k + 4h, which simplify to y = 4x - 14 and y = -4x + 2.

Step 5: Verify with Graphing Utility

Plot the given hyperbola and the calculated asymptotes using a graphing utility to verify the results. The hyperbola should touch the asymptotes at the vertices and the foci should fall on the vertical axis of the hyperbola.

Key Concepts

Coordinates of Center

Understanding the center of a hyperbola is crucial as it is the mid-point from which the rest of its geometric features are determined. The center is found at the coordinates \textbf{(h, k)}. In our case, given the equation \(\frac{(y+6)^{2}}{1}-\frac{(x-2)^{2}}{\frac{1}{16}}=1\), we deduce the center by looking at the transformations of the variables x and y. Here, the equation indicates a horizontal shift to the right by 2 and a vertical shift down by 6, hence yielding the center coordinates at \((2, -6)\).

Vertices and Foci of Hyperbola

The vertices of a hyperbola are points where it intersects its axis of symmetry and are located 'a' units from the center. In this example, the vertices are at coordinates \((2, -5)\) and \((2, -7)\), determined by adding and subtracting 'a', which equals 1 here, to the y-coordinate of the center.

The foci are points inside the hyperbola, through which the difference of distances from any point on the hyperbola to the foci is constant. They lie 'c' units from the center, with \c = \sqrt{a^2 + b^2}\. The calculation gives us \c = \sqrt{17}/4\, placing the foci at \((2, -6 \pm \sqrt{17}/4)\). This shows that the foci are on a vertical line through the center, as our hyperbola is vertical based on the given equation.

Asymptotes of Hyperbola

Asymptotes guide us on how the arms of the hyperbola behave at infinity – they approach these lines but never intersect them. To find the asymptotes for our hyperbola, we use the slopes \(\pm a/b\). With \(a = 1\) and \(b = 1/4\), the slopes are \(\pm 4\).

Creating the equations for these lines through the center, we have \(y = 4x - 14\) and \(y = -4x + 2\), determining two straight lines that dictate the direction of the opening of the hyperbola. Remember, because the hyperbola opens along the vertical axis, these asymptotes, having non-zero slopes, indicate the direction of the branches is diagonal and cross at the center.

Graphing Utility Verification

It's always a good idea to check your work, and graphing utilities are invaluable tools for this. After finding theoretical values for the center, vertices, foci, and asymptotes, use graphing software to visualize the hyperbola. When you plot \(\frac{(y+6)^{2}}{1}-\frac{(x-2)^{2}}{\frac{1}{16}}=1\), you should see that the hyperbola touches the lines of the asymptotes at the vertices.

The graph should also indicate the foci on the vertical axis passing through the center, confirming the accuracy of our calculations. This practical verification helps ensure that we have comprehended all parts of the hyperbola and its characteristics correctly.