Problem 19 - 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. $$4 y^{2}-9 x^{2}=144$$

Step-by-Step Solution

Answer

Vertices: (0, ±6); Foci: (0, ±2√13); Asymptotes: y=±3/2x; Transverse axis length: 12.

Step 1: Convert to Standard Form

Start by writing the given equation in standard form. Divide the entire equation by 144:\[ \frac{4y^2}{144} - \frac{9x^2}{144} = \frac{144}{144} \]Simplify to get:\[ \frac{y^2}{36} - \frac{x^2}{16} = 1 \]This is the standard form of a vertical hyperbola centered at the origin \((0,0)\).

Step 2: Identify Key Hyperbola Components

Given the standard form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), identify \(a^2 = 36\) and \(b^2 = 16\). Thus, \(a = 6\) and \(b = 4\). The vertices are located at \((0, \pm a) = (0, \pm 6)\). To find the foci, calculate \(c = \sqrt{a^2 + b^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}\). The foci are at \((0, \pm 2\sqrt{13})\).

Step 3: Find Asymptotes

For a vertical hyperbola, the asymptotes are given by the equations:\[ y = \pm \frac{a}{b} x \] Substituting \(a = 6\) and \(b = 4\), the equations for the asymptotes are:\[ y = \pm \frac{6}{4} x = \pm \frac{3}{2} x \]

Step 4: Determine Transverse Axis Length

The length of the transverse axis is given by \(2a\). Since \(a = 6\), the transverse axis is \(2 \times 6 = 12\).

Step 5: Sketch the Graph

Plot the hyperbola by sketching its vertices, locating the foci, and drawing the asymptotes from the center as guides. The vertices, foci, and asymptotes guide the shape, illustrating a vertical hyperbola open up and down.

Key Concepts

Vertices of a Hyperbola

In the context of a hyperbola, the vertices are key points that define its shape. For the hyperbola given by the equation \( \frac{y^2}{36} - \frac{x^2}{16} = 1 \), we know it's a vertical hyperbola centered at the origin (0,0). The standard form of a vertical hyperbola \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \) is used to find the location of the vertices.The value \(a\) is crucial in determining the position of the vertices. Here, \(a^2 = 36\), which means \(a = 6\). This tells us the vertices are located at

• (0, a) which is (0, 6), and
• (0, -a) which is (0, -6).

These points signify the positions on the y-axis where the hyperbola intercepts, forming the top and bottom of its vertical shape.Understanding vertices helps in sketching the basic framework of the hyperbola, giving you a clear visualization of its overall structure.

Foci of a Hyperbola

The foci of a hyperbola are points located along the hyperbola's axis of symmetry, further apart than the vertices. They play a key role in defining the hyperbola's properties and location.To find the foci of our hyperbola, we use the relationship:\[c = \sqrt{a^2 + b^2}\]Where \(a^2 = 36\) and \(b^2 = 16\), giving us:\(c = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}\).For a vertical hyperbola like \( \frac{y^2}{36} - \frac{x^2}{16} = 1 \), the calculated foci are at:

• (0, c) which is (0, 2\sqrt{13}), and
• (0, -c) which is (0, -2\sqrt{13}).

These points demonstrate the real physical 'stretch' of the hyperbola, extending further than the vertices do. The foci direct the characteristic nature and wideness of the hyperbola, as every point on the hyperbola is such that the difference of the distances to the foci is constant.

Asymptotes of a Hyperbola

Asymptotes are lines that a hyperbola approaches but never actually touches. They act as a guide to the hyperbola's shape, showing the direction and inclination of its branches.For a vertical hyperbola like \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the asymptotes are given by the equations:\[ y = \pm \frac{a}{b} x \]Substituting \(a = 6\) and \(b = 4\), we find:\[ y = \pm \frac{6}{4} x = \pm \frac{3}{2} x \]These are the asymptote lines \(y = \frac{3}{2} x\) and \(y = -\frac{3}{2} x\) that crisscross through the origin.

• They provide a vital guide for sketching the hyperbola, as its branches will get increasingly closer to these lines without crossing them.
• Asymptotes highlight that as you move away from the center, the branches progressively "open up" within these boundaries.

Understanding the asymptotes helps to graph the hyperbola accurately and understand its expansive nature.