Problem 24 - Algebra and Trigonometry

Question

use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. $$ 16 y^{2}-9 x^{2}=144 $$

Step-by-Step Solution

Answer

The vertices are (0,3) and (0,-3), the asymptotes are \(y = \frac{3x}{4}\) and \(y = -\frac{3x}{4}\), and the foci are at (0,5) and (0,-5). With this, the hyperbola can be plotted on a graph reflecting the problem's conditions.

Step 1: Identifying the Standard Form

First, rewrite the equation so that it's in the standard form for hyperbolas \((y-k)^2/a^2 - (x-h)^2/b^2 = 1\). Dividing the given equation by 144 gives the standard form: \((y-0)^2/9 - (x-0)^2/16 = 1\) which implies \(h=0\), \(k=0\), \(a^2=9\), and \(b^2=16\).

Step 2: Finding the Vertices, Asymptotes and Foci

The vertices are located at \((h, k \pm a)\), or \((0, \pm \sqrt{9}) = (0, \pm 3)\). The line equations for the asymptotes are \((y-k) = \pm (a/b)(x-h)\) or \(y = \pm (3/4)x\). The foci are at \((h, k \pm \sqrt{a^2+b^2})\), thus coordinates of the foci are given by: \((0, \pm \sqrt{9+16}) = (0, \pm 5)\).

Step 3: Plotting the Graph

Draw the axes on a graph paper and mark the vertices and the foci. Draw a rectangle based on a and b distances from the center and the asymptotes. The intersections of the asymptotes and the rectangle provide points on the hyperbola. Draw the hyperbola, keeping in mind that it has to approach the asymptotes.

Key Concepts

Vertices in Hyperbolas

In a hyperbola, the vertices are crucial points that lie on the hyperbola's transverse axis. These points help in defining the shape and the overall look of the hyperbola. For a hyperbola given in the standard form of \[ (y-k)^2/a^2 - (x-h)^2/b^2 = 1, \]the vertices are calculated by adding and subtracting the value \( a \) from the \( k \) coordinate, all while the \( h \) coordinate remains constant. This results in the vertices being located at:

• \((h, k + a)\)
• \((h, k - a)\)

in specific terms: \((0, 3)\) and \((0, -3)\) when \( a = 3 \).
These vertices not only help in graphing the hyperbola but also in understanding how the hyperbola will stretch along its axis. By marking these on a graph, you get a clear indication of where the hyperbola will be located, and how wide it will appear across the transverse axis.

Understanding Asymptotes

Asymptotes are essential in guiding the hyperbola's shape by indicating the direction the hyperbola will take as it moves away from the center. They are straight lines that the hyperbola approaches but never actually touches.For a hyperbola in the form \[ (y-k)^2/a^2 - (x-h)^2/b^2 = 1, \] the equations of the asymptotes are calculated as:

• \( (y-k) = \pm (a/b)(x-h) \)

In our example, this becomes:

• \( y = \pm (3/4)x \)

This equation illustrates how steeply the branches of the hyperbola will open. The slope, \( 3/4 \), is determined by the ratio of \( a \) to \( b \).
When drawing the hyperbola, use these asymptotes as imaginary guideline lines to sketch the curve. Asymptotes help ensure that the hyperbola's branches trend outwards correctly, making them vital to plotting the full hyperbolic curve correctly.

Foci of a Hyperbola

The foci are interior points essential to a hyperbola's definition. They exist such that the difference in distances from any point on the hyperbola to the two foci is always constant. These points are located further along the transverse axis beyond the vertices.For the standard form:\[ (y-k)^2/a^2 - (x-h)^2/b^2 = 1, \]the foci are found using the formula:

• \((h, k \pm \sqrt{a^2 + b^2})\)

For our given example, the foci are positioned at:

• \((0, \pm \sqrt{9 + 16})\)
• \((0, \pm 5)\)

These foci points are crucial as they help to determine the magnitude of separation the branches of the hyperbola have. Making these points visible on your graph aids in comprehensively understanding how the hyperbola is configured spatially relative to the center.