Problem 18 - College Algebra

Question

Use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. $$\frac{y^{2}}{25}-\frac{x^{2}}{64}=1$$

Step-by-Step Solution

Answer

The hyperbola is centered at the origin, with vertices at (0,5) and (0,-5). The foci are at (0, ±√89). The equations of the asymptotes are \(y = ± \frac{5}{8}*x\).

Step 1: Determination of the center, vertices, and a and b

First, identify the center of the hyperbola, which in this case is at the origin due to the nature of the equation given. Then, determine a and b for this hyperbola. Here \(a = \sqrt{25} = 5\) and \(b = \sqrt{64} = 8\). So the vertices of the hyperbola will be at (0, 5) and (0, -5) because it is a vertical hyperbola.

Step 2: Calculation of the foci

Next, calculate the foci. Use the formula \(c = \sqrt{a^2 + b^2}\). Here, \(c = \sqrt{5^2 + 8^2} = \sqrt{25 + 64} = \sqrt{89}\). Therefore, the foci will be at (0, ±√89).

Step 3: Equation of the asymptotes

Finally, determine the equations of the asymptotes. For a hyperbola centered at the origin, the asymptotes are given by the formula \(y = ± \frac{a}{b}*x\). Here, the equations of the asymptotes will be \(y = ± \frac{5}{8}*x\).

Key Concepts

Asymptotes of a Hyperbola

To fully grasp the concept of hyperbolas, understanding asymptotes is a critical piece. An asymptote of a hyperbola is a line that the hyperbola approaches but never actually reaches. These are imaginary lines that provide a skeleton to which the hyperbola adheres, shaping its open-ended curves. For the hyperbola given by the equation \( \frac{y^2}{25} - \frac{x^2}{64} = 1 \), the asymptotes play a pivotal role in graphing.

As the hyperbola is centered at the origin, its asymptotes pass through the center. They offer a clear guide for graphing by outlining the direction in which the hyperbola's branches extend as they move further away from the center. The standard formula for the equations of these asymptotes is \( y = \pm \frac{a}{b}*x \) when the hyperbola is centered at the origin. In this case, with values \( a = 5 \) and \( b = 8 \) incorporated, the equations simplify to \( y = \pm \frac{5}{8}*x \), depicting that they will have a slope of \( \pm \frac{5}{8} \) leading away from the origin in both positive and negative directions of the y-axis.

Hyperbola Vertices

The vertices of a hyperbola hold significance as they are the points that give us the closest approach of the hyperbola to its center. For the vertical hyperbola \( \frac{y^2}{25} - \frac{x^2}{64} = 1 \), these vertices are found along the y-axis given that the y-term is positive. To determine the location of the vertices, we look for the value of \( a \) from the equation, which is \( \sqrt{25} = 5 \) in this case.

Placing the value of \( a \) in coordinates relative to the center (which is at the origin (0,0)), we obtain the vertices (0, 5) and (0, -5). These points represent the 'tips' of the hyperbola's arches where the curve is at its sharpest. By identifying these vertices, graphing the hyperbola becomes a matter of plotting these points and using the asymptotes to guide the curve asymptotically away from these key locations.

Hyperbola Foci

Going beyond the vertices, the foci of a hyperbola are points inside each curve that are essential in defining its shape. The foci are located along the principal axis, the same vertical line where the vertices reside for a vertically oriented hyperbola. To find the position of the foci for the given equation \( \frac{y^2}{25} - \frac{x^2}{64} = 1 \) we use \( c \) from the formula \( c = \sqrt{a^2 + b^2} \), which turns out to be \( \sqrt{89} \) for our example.

Understanding that the foci (0, ±√89) lie outside the vertices on the y-axis helps to visualize the hyperbola's extent beyond the vertices themselves. A common error in graphing is mistaking the vertices for the foci. However, remembering that the foci align with the innermost part of the hyperbola's 'bow' and are distant from the center by \( c \), further than the distance the vertices are (which is \( a \) units from the center), assists in accurate visualization and graphing of hyperbolas. The foci, due to their unique geometric location, are crucial points that affect the overall form and equation of the hyperbola.