Problem 10 - 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. $$\frac{y^{2}}{9}-\frac{x^{2}}{16}=1$$

Step-by-Step Solution

Answer

Vertices: (0, 3) and (0, -3), Foci: (0, 5) and (0, -5), Asymptotes: y = ±3/4x, Transverse axis length: 6.

Step 1: Identify the Standard Form

The given equation is \(\frac{y^{2}}{9}-\frac{x^{2}}{16}=1\). Compare this with the standard form of a vertical hyperbola \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). We identify \(a^2 = 9\) and \(b^2 = 16\). This gives \(a = 3\) and \(b = 4\).

Step 2: Calculate Vertices

For a vertical hyperbola, the vertices are located at \((0, \pm a)\). Since \(a = 3\), the vertices are \((0, 3)\) and \((0, -3)\).

Step 3: Calculate Foci

The foci of a vertical hyperbola are \((0, \pm c)\) where \(c = \sqrt{a^2 + b^2}\). Calculate \(c = \sqrt{9 + 16} = \sqrt{25} = 5\). Thus, the foci are \((0, 5)\) and \((0, -5)\).

Step 4: Determine Asymptotes

For a vertical hyperbola \(\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1\), the asymptotes have equations \(y = \pm \frac{a}{b}x\). Here, \(a = 3\) and \(b = 4\), so the asymptotes are \(y = \pm \frac{3}{4}x\).

Step 5: Determine Length of Transverse Axis

The transverse axis length for a vertical hyperbola is \(2a\). With \(a = 3\), the length is \(2 \times 3 = 6\).

Step 6: Sketch the Hyperbola

Plot the center at the origin \((0, 0)\), mark vertices at \((0, 3)\) and \((0, -3)\), plot the foci at \((0, 5)\) and \((0, -5)\), and draw the asymptotes \(y = \pm \frac{3}{4}x\). Sketch the hyperbola opening upwards and downwards, staying close to the asymptotes as it extends.

Key Concepts

Vertices of Hyperbolas

Vertices are specific points on a hyperbola that help in understanding its orientation and shape. For the given equation \(\frac{y^{2}}{9}-\frac{x^{2}}{16}=1\), it is a vertical hyperbola. This is known because the \(y^2\) term comes first.

The vertices for vertical hyperbolas are directly related to the variable \(a\), where \(a^2\) is under the \(y^2\) term. Here, \(a^2 = 9\) and therefore \(a = 3\).

• Place the vertices at \((0, a)\) and \((0, -a)\).
• For our specific hyperbola, these points are \((0, 3)\) and \((0, -3)\).

These points are crucial because they indicate where the hyperbola passes through the transverse axis and mark the extent of its opening.

Foci of Hyperbolas

The foci of the hyperbola are another set of significant points located along the transverse axis, beyond the vertices. They are essential because they determine the hyperbola's eccentricity and provide insight into its shape.
In the equation \(\frac{y^{2}}{9}-\frac{x^{2}}{16}=1\), the foci are located at \((0, \pm c)\), where \(c\) is given by \(c = \sqrt{a^2 + b^2}\).

• For the given hyperbola, \(c = \sqrt{9 + 16} = \sqrt{25} = 5\).
• Hence, the foci are situated at \((0, 5)\) and \((0, -5)\).

These points are always positioned inside the hyperbola's arms, and the hyperbola tends to wrap around these focus points.

Asymptotes of Hyperbolas

Asymptotes are lines that a hyperbola approaches but never touches. They guide the shape of the hyperbola, helping us draw it accurately.

For vertical hyperbolas that follow the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the asymptotes can be expressed as \(y = \pm \frac{a}{b}x\). From the equation \(\frac{y^{2}}{9}-\frac{x^{2}}{16}=1\) we have:

• \(a = 3\) and \(b = 4\)
• Thus, the asymptote equations are \(y = \pm \frac{3}{4}x\).

These lines intersect at the center of the hyperbola and slope outwards, forming a cross pattern that aligns with the hyperbola's opening. The hyperbola will monotonously approach and trail this asymptotic path, particularly as it extends away from the center.

Transverse Axis Length

The transverse axis length is a simple but vital feature of hyperbolas. It represents the full length between the vertices through the center.
For a vertical hyperbola such as \(\frac{y^{2}}{9}-\frac{x^{2}}{16}=1\), the transverse axis is vertical, since the main term \(y^2\) is in front.

• The length is determined by \(2a\).
• Given \(a = 3\), the length of the transverse axis would be \(2 \times 3 = 6\).

Understanding this distance helps in sketching the hyperbola accurately and acts as a guide to how broad the central part of the hyperbola is.

Graphing Hyperbolas

Graphing a hyperbola can initially seem daunting but becomes straightforward if done step by step. Follow these tips:
1. Start by plotting the center, which is at \((0, 0)\) for the hyperbola \(\frac{y^{2}}{9}-\frac{x^{2}}{16}=1\).
2. Next, mark the vertices at \((0, 3)\) and \((0, -3)\), as these show where the hyperbola crosses the transverse axis.
3. Plot the foci at \((0, 5)\) and \((0, -5)\), which fall along the transverse axis.
4. Draw the asymptotes using the equations \(y = \pm \frac{3}{4}x\). These serve as guidelines for the hyperbola's shape.
5. Finally, sketch the two branches of the hyperbola that open upwards and downwards, following close to the asymptotes as they extend. Piecing everything together properly will give you a visually accurate picture of the hyperbola's architecture, illustrating both its symmetry and elegance.