Problem 16 - Precalculus: Mathematics for Calculus

Question

An equation of a hyperbola is given. (a) Find the vertices, foci, and asymptotes of the hyperbola. (b) Determine the length of the transverse axis. (c) Sketch a graph of the hyperbola. $$\frac{x^{2}}{16}-\frac{y^{2}}{12}=1$$

Step-by-Step Solution

Answer

Vertices: (4, 0), (-4, 0); Foci: (~5.29, 0), (~-5.29, 0); Asymptotes: y = ±0.865x; Transverse axis length: 8.

Step 1: Identify the form of the hyperbola

The given equation is \( \frac{x^2}{16} - \frac{y^2}{12} = 1 \). This matches the standard form of a hyperbola: \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), where \( a^2 = 16 \) and \( b^2 = 12 \). Identify \( a = 4 \) and \( b \approx 3.46 \). This is a horizontally oriented hyperbola.

Step 2: Find the vertices

The vertices of the hyperbola are determined from \( a \). For a horizontal hyperbola, the vertices are \( (\pm a, 0) \). Thus, the vertices are \( (\pm 4, 0) \), giving us the points \( (4, 0) \) and \( (-4, 0) \).

Step 3: Locate the foci

The distance to each focus (\( c \)) is found using \( c = \sqrt{a^2 + b^2} \). So, \( c = \sqrt{16 + 12} = \sqrt{28} \approx 5.29 \). The foci are at \( (\pm c, 0) \), giving points approximately at \( (5.29, 0) \) and \( (-5.29, 0) \).

Step 4: Determine the asymptotes

The equations for the asymptotes of a hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) are \( y = \pm \frac{b}{a} x \). Thus, the asymptotes are \( y = \pm \frac{3.46}{4} x \) or approximately \( y = \pm 0.865 x \).

Step 5: Length of the transverse axis

The transverse axis of a hyperbola is the distance between the vertices, calculated as \( 2a \). With \( a = 4 \), the transverse axis has a length of \( 2 \times 4 = 8 \).

Step 6: Sketch the hyperbola

To sketch the hyperbola, plot the vertices \( (4, 0) \) and \( (-4, 0) \), the foci around \( (5.29, 0) \) and \( (-5.29, 0) \), and draw the asymptotes \( y = \pm 0.865x \). The hyperbola opens horizontally and approaches but never touches the asymptotes.

Key Concepts

Vertices of Hyperbola

The vertices of a hyperbola are crucial points that lie at the intersection of the hyperbola with its transverse axis. For the given hyperbola with equation \( \frac{x^2}{16} - \frac{y^2}{12} = 1 \), the vertices help define the shape and orientation initially. Since this is a horizontally oriented hyperbola, the vertices are calculated using \( a \), derived from the equation's denominator where the variable \( x \) is involved, which is 16. Knowing \( a^2 = 16 \), we get \( a = 4 \).
Thus, the vertices of the hyperbola are located at \((\pm a, 0)\), or more specifically, at the points \((4, 0)\) and \((-4, 0)\). These points are key to sketching the hyperbola and understanding its orientation relative to its center at the origin. By identifying these points, students can better grasp the symmetry and expanse of hyperbolas as they relate to the transverse axis.

Foci of Hyperbola

Foci are integral to understanding hyperbolas, as they determine how the curves bend around an axis. The foci for a hyperbola can be found using the formula \( c = \sqrt{a^2 + b^2} \). From our original hyperbola equation, we have \( a^2 = 16 \) and \( b^2 = 12 \), leading to the calculation:

• \( c = \sqrt{16 + 12} = \sqrt{28} \approx 5.29 \)

Using this result, the foci are located at \((\pm c, 0)\), or approximately at the positions \((5.29, 0)\) and \((-5.29, 0)\).
These focus points are not just central to the geometry of the hyperbola; they reveal how the hyperbola extends into infinity. Even though they don't lie on the hyperbola itself, the focus (plural: foci) helps define the curve's width and how it 'opens' outwards.

Transverse Axis

The transverse axis of a hyperbola is always aligned along the direction of the vertices and serves as the main axis through which the hyperbola is considered to 'open.' It's a line segment connecting the vertices and its length is particularly significant in understanding the hyperbola's span.

• For a horizontal hyperbola, the transverse axis runs left to right crossing the center at the origin.
• The length is calculated as \( 2a \), with \( a = 4 \), yielding a total length of \( 8 \).

Understanding this axis offers a visual aid in sketching and interpreting the hyperbola's shape. It is the major line that defines the width from one vertex to the other. As such, it's important in solving problems involving intersections and other geometric properties of hyperbolas.

Asymptotes of Hyperbola

Asymptotes for hyperbolas act as boundary lines the curves tend to approach but never actually meet. They help shape the 'opening' of the hyperbola and are crucial for sketching accurate representations, especially over large distances where the curve appears nearly straight.l
For the equation \( \frac{x^2}{16} - \frac{y^2}{12} = 1 \), the asymptote formula for a hyperbola is \( y = \pm \frac{b}{a} x \). With \( b = 3.46 \) and \( a = 4 \), the asymptote equations become:

• \( y = \pm \frac{3.46}{4} x \)
• approximately \( y = \pm 0.865 x \)

These lines not only aid in sketching the hyperbola but also provide deep insight into the hyperbola's expansiveness as it moves further from the origin. Even though they don't touch the hyperbola, the relationship informs about symmetry and curvature as one traces along the hyperbola's path.