Problem 16 - Algebra and Trigonometry

Question

use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. $$ \frac{x^{2}}{144}-\frac{y^{2}}{81}=1 $$

Step-by-Step Solution

Answer

The vertices of the hyperbola are located at (±12,0). The foci are located at (±15,0). The equations for the asymptotes are \(y = ± \frac{3}{4}x\).

Step 1: Determine Center, Vertices, and Foci

The center of the hyperbola is at the origin (0,0). The values of a and b are the square roots of the denominators of x and y in the equation, respectively. Thus, a = \(\sqrt{144}\) = 12 and b = \(\sqrt{81}\) = 9. The vertices are at distances 'a' from the center along the x-axis since 'a' is along the x-axis. Hence, vertices are (±12, 0). The distance from the center to the foci (c) is determined by the equation \(c = \sqrt{a^2 + b^2}\). By substitution, c = \(\sqrt{12^2 + 9^2}\) = 15. So, the foci are at (±15, 0).

Step 2: Find the Asymptotes

The equation of the asymptotes for a hyperbola in this orientation is \(y = ± \frac{b}{a}x\). Substituting our values, we get \(y = ± \frac{9}{12}x\) or \(y = ± \frac{3}{4}x\)

Step 3: Plot the Hyperbola

Start by plotting the center at (0,0). Then mark the vertices at (±12, 0) and the foci at (±15, 0). Plot the asymptotes, these are the lines \(y = ± \frac{3}{4}x\). The asymptotes give the 'opening' of the hyperbola. Draw the smooth curve of the hyperbola opening toward the vertices and approaching the asymptotes as the lines move away from the center.

Key Concepts

Vertices

When talking about hyperbolas, the vertices are essential reference points. Vertices are the points where the hyperbola intersects the transverse axis. In simpler terms, they are the "tips" of the hyperbola.

• To find the vertices of a hyperbola given by the equation \( \frac{x^2}{144} - \frac{y^2}{81} = 1 \), you need the value of \(a\), which corresponds to the distance from the center to each vertex.
• Here, \(a = \sqrt{144} = 12\).
• This means the vertices are at (±12, 0) along the x-axis.

Located symmetrically from the center, these points help in sketching the general shape and direction of the hyperbola. Understanding vertices is crucial for plotting the hyperbola correctly.

Asymptotes

Asymptotes of a hyperbola are invisible guide lines that the hyperbola approaches but never touches. They help to determine the 'spread' of the hyperbola.

• For the hyperbola \( \frac{x^2}{144} - \frac{y^2}{81} = 1 \), the equations of the asymptotes are determined by the formula \( y = \pm \frac{b}{a}x \).
• Plugging in the values, you get \( b = 9\) and \( a = 12\), which makes the asymptotes' equations \( y = \pm \frac{3}{4}x \).

These diagonal lines through the center (0,0) set the direction for the hyperbola's arms. They visually 'frame' the overall path and help avoid errors when drawing the curve.

Foci

The foci of a hyperbola are two fixed points that are used to define and construct the hyperbola. These are the points that are most central to how the hyperbola is shaped.

• The formula to find the distance from the center to each focus is \( c = \sqrt{a^2 + b^2} \).
• For the equation \( \frac{x^2}{144} - \frac{y^2}{81} = 1 \), we calculate \(c\) as \( \sqrt{12^2 + 9^2} = 15 \).
• This places the foci at \( (\pm 15, 0) \).

The foci are crucial since every point on the hyperbola has a characteristic relationship with them. Specifically, the absolute difference of distances from any point on the hyperbola to the two foci is constant. Knowing where the foci are can dramatically improve understanding of the hyperbola's form.

Graphing Hyperbolas

Graphing hyperbolas is about putting together all the key elements found in its equation. Here is a simple breakdown to help graph the hyperbola \( \frac{x^2}{144} - \frac{y^2}{81} = 1 \).

• First, identify the center, which is at (0,0) for standard equations.
• Next, mark vertices at (±12, 0), as they give a general direction horizontally.
• Then, draw the asymptotes, \( y = \pm \frac{3}{4}x \). They help align the drawing.
• Locate the foci at (±15, 0), which influences the curve's tightness.
• Finally, sketch the actual hyperbola by drawing smooth curves going through the vertices and asymptotes.

Often, the hyperbola will appear as two mirrored arcs, opening away from the center. By plotting these elements correctly, you're delicately weaving the hyperbola's shape, ensuring precision and accuracy.