Problem 62 - Precalculus: Functions and Graphs

Question

Find an equation for the indicated part of the hyperbola. Left halves of the branches of \(\frac{y^{2}}{36}-\frac{x^{2}}{16}=1\)

Step-by-Step Solution

Answer

The equation is \( x = -\sqrt{ \frac{y^{2}}{36} - 1} \cdot 4 \) for \( y^{2} \geq 36 \).

Step 1: Understanding the Standard Form

The standard form of a hyperbola oriented vertically is \( \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \). The given equation \( \frac{y^{2}}{36} - \frac{x^{2}}{16} = 1 \) is already in this form, where \(a^{2} = 36\) and \(b^{2} = 16\).

Step 2: Determine Half of the Hyperbola to Focus

Since we are interested in the left halves of the hyperbola's branches, we need to consider the parts of the graph where \( x \leq 0 \).

Step 3: Rewrite Equations for Left Halves

Since \( y^{2}/36 - x^{2}/16 = 1 \) represents two branches vertically, solve for \(x\):\[ x = \pm \sqrt{ \frac{y^{2}}{36} - 1} \cdot 4.\] We select the left half by choosing \( x = -\sqrt{ \frac{y^{2}}{36} - 1} \cdot 4 \) where \( y \in \mathbb{R} \).

Step 4: Combine Results

Therefore, the equation for the left halves of the hyperbola's branches is \( x = -\sqrt{ \frac{y^{2}}{36} - 1} \cdot 4 \) with all values of \( y \) where this makes \( x \) real (i.e., where \( y^{2} \geq 36 \)).

Key Concepts

Standard Form of Hyperbola

Hyperbolas are fascinating shapes in mathematics, known for their unique two-branch structure. The standard form of a hyperbola is different depending on its orientation—there is a standard form for vertically and horizontally oriented hyperbolas. For a vertically oriented hyperbola, the standard form is given by:
\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]
In this form:

• \(a^2\) represents the squared value of the semi-major axis length vertically.
• \(b^2\) is associated with the semi-minor axis horizontally.

In our original exercise equation, \(\frac{y^2}{36} - \frac{x^2}{16} = 1\), we see it is already in the standard vertical orientation form, with \(a^2 = 36\) and \(b^2 = 16\). From \(a^2 = 36\), we can deduce that \(a = 6\), while \(b^2 = 16\) gives us \(b = 4\). This means the hyperbola opens up and down along the y-axis.

Equation of Hyperbola

The equation of a hyperbola can tell us a lot about its shape and orientation. By understanding the orientation either vertical or horizontal, you can easily determine which axis the hyperbola opens along. In terms of components:
- The minus sign in \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) signals a hyperbola.- The terms \(\frac{y^2}{a^2}\) and \(\frac{x^2}{b^2}\) will lead us to its orientation.
For the equation \(\frac{y^2}{36} - \frac{x^2}{16} = 1\), because the \(y^2\) term comes first, and is positive, the hyperbola is vertical.
These terms also help us to determine the hyperbola’s asymptotes, which are lines that the hyperbola approaches but never touches. The equations of the asymptotes can be found as \(y = \pm \frac{a}{b}x\), or specifically here \(y = \pm \frac{6}{4}x\).

Graphing Hyperbolas

Graphing a hyperbola involves understanding its shape, orientation, and features like asymptotes and vertices. For the given equation \(\frac{y^2}{36} - \frac{x^2}{16} = 1\), we can extract valuable insights to graph it effectively:
- **Vertices:** These are points where the hyperbola intersects the y-axis. Here, the vertices are found by setting \(x = 0\) in the equation, leading to \(y = \pm 6\).- **Center:** Found at the origin \((0,0)\) since there are no values added/subtracted to \(x\) or \(y\).- **Asymptotes:** These include lines \(y = \pm \frac{6}{4}x\) which guide the graph and can be used as reference lines that the branches of the hyperbola approach as they extend outward.
To focus on the left halves of the branches, choosing \(x = -\sqrt{ \frac{y^2}{36} - 1} \cdot 4\) helps in isolating one half of the graph. This means visually reducing our focus to sections where \(x\) remains less than or equal to zero. This is crucial when interested only in one section of a hyperbola.