Problem 16 - Precalculus: Mathematical for Calculus

Question

Find the vertices, foci, and asymptotes of the hyperbola, and sketch its graph. $$9 x^{2}-16 y^{2}=1$$

Step-by-Step Solution

Answer

Vertices: \( (\pm \frac{1}{3}, 0) \); Foci: \( (\pm \frac{5}{12}, 0) \); Asymptotes: \( y = \pm \frac{3}{4}x \).

Step 1: Identify Standard Form

First, we need to write the given equation of the hyperbola in its standard form: \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). The given equation is \( 9x^2 - 16y^2 = 1 \), which can be rewritten as \( \frac{x^2}{\frac{1}{9}} - \frac{y^2}{\frac{1}{16}} = 1 \). Thus, the standard form is \( \frac{x^2}{\frac{1}{9}} - \frac{y^2}{\frac{1}{16}} = 1 \).

Step 2: Determine Values of a and b

From the standard form, we have \( a^2 = \frac{1}{9} \) and \( b^2 = \frac{1}{16} \). Consequently, \( a = \frac{1}{3} \) and \( b = \frac{1}{4} \).

Step 3: Identify the Vertices

The vertices of a hyperbola in the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) with the transverse axis along the x-axis are \( (\pm{a}, 0) \). Thus, the vertices are \( (\pm{\frac{1}{3}}, 0) \).

Step 4: Calculate the Foci

The formula for the foci of a hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is derived from \( c^2 = a^2 + b^2 \). Compute \( c^2 = \frac{1}{9} + \frac{1}{16} = \frac{25}{144} \), so \( c = \frac{5}{12} \). The foci are \( (\pm{\frac{5}{12}}, 0) \).

Step 5: Find the Asymptotes

For a hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the asymptotes are given by the equations \( y = \pm \frac{b}{a}x \). Thus, \( y = \pm \frac{\frac{1}{4}}{\frac{1}{3}}x = \pm \frac{3}{4}x \).

Step 6: Sketch the Graph

To sketch the hyperbola, first plot the vertices at \( (\pm \frac{1}{3}, 0) \), then sketch the asymptotes as lines passing through the origin with slopes \( \pm \frac{3}{4} \). Use the vertices and asymptotes to draw the two branches of the hyperbola symmetrically about the x-axis.

Key Concepts

Vertices of a Hyperbola

The vertices of a hyperbola are crucial points where each branch of the hyperbola is closest to the center. By understanding the position of the vertices, we can determine the shape and orientation of the hyperbola. For a hyperbola in standard form \[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,\]the vertices lie along the x-axis, assuming the transverse axis is parallel to the x-axis. The distance from the center to each vertex along the transverse axis is represented by the value of \( a \).
In our specific example, with \( a = \frac{1}{3} \), the vertices are located at:

• \( \left( \frac{1}{3}, 0 \right) \)
• \( \left(-\frac{1}{3}, 0 \right) \)

The vertices guide us in sketching the primary structure of the hyperbola and set the bounds for the hyperbola's central opening. Knowing these points is essential when graphing, as they show the widest points of the curves in relation to the center.

Foci of a Hyperbola

The foci are another set of important points for hyperbolas. These points help define the hyperbola’s intrinsic property – that the difference in distances from any point on the hyperbola to the foci is constant. This characteristic differentiates hyperbolas from other conic sections like ellipses.
For hyperbolas in the standard horizontal form \[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,\]we determine the foci using the relationship \( c^2 = a^2 + b^2 \). Hence, \( c \) is the distance from the center to each focus along the transverse axis.
In our example, with calculated \( c = \frac{5}{12} \), the foci are:

• \(\left( \frac{5}{12}, 0 \right)\)
• \(\left(-\frac{5}{12}, 0 \right)\)

These foci lie further out along the x-axis compared to the vertices, indicating that the branches of the hyperbola stretch toward these points.

Asymptotes of a Hyperbola

Asymptotes are straight lines that the hyperbola approaches as \( x \) or \( y \) goes to infinity. They provide a guide that indicates the direction of the curves on each branch of the hyperbola. These lines do not intersect the hyperbola but act as a boundary that the hyperbola never crosses.
For hyperbolas in the standard form \[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, \]the asymptotes are described by \( y = \pm \frac{b}{a} x \). This relationship is derived from evaluating key points at infinity.In our exercise, since we have \( a = \frac{1}{3} \) and \( b = \frac{1}{4} \), the asymptotes become:

• \( y = \frac{3}{4}x \)
• \( y = -\frac{3}{4}x \)

These asymptotes cross at the hyperbola's center and visually constrain the path of each branch. They are indispensable in sketching, helping to make the hyperbola appear symmetric and correctly oriented.