Problem 15 - Precalculus: Mathematical for Calculus

Question

Find the vertices, foci, and asymptotes of the hyperbola, and sketch its graph. $$4 y^{2}-x^{2}=1$$

Step-by-Step Solution

Answer

Vertices: (0, ±1/2); Foci: (0, ±√5/2); Asymptotes: y = ±1/2x.

Step 1: Rewrite the Equation in Standard Form

The given equation is \(4y^2 - x^2 = 1\). Begin by rewriting it in the standard form of a hyperbola, which is \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). Divide every term by 1 to maintain equality:\[\frac{4y^2}{1} - \frac{x^2}{1} = \frac{1}{1}\]This simplifies to \(\frac{y^2}{\frac{1}{4}} - \frac{x^2}{1} = 1\), so \(a^2 = \frac{1}{4}\) and \(b^2 = 1\).

Step 2: Identify the Vertices

Since the hyperbola is of the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the vertices are located at \((0, \pm a)\). Calculate \(a = \sqrt{\frac{1}{4}} = \frac{1}{2}\). Thus, the vertices are \((0, \frac{1}{2})\) and \((0, -\frac{1}{2})\).

Step 3: Find the Foci

The foci of the hyperbola are located at \((0, \pm c)\) for a vertical hyperbola, where \(c^2 = a^2 + b^2\). Calculate \(c\):\[c^2 = \frac{1}{4} + 1 = \frac{5}{4}\]\(c = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}\).Therefore, the foci are \((0, \frac{\sqrt{5}}{2})\) and \((0, -\frac{\sqrt{5}}{2})\).

Step 4: Determine the Asymptotes

For a vertical hyperbola, the asymptotes are given by the equations \(y = \pm \frac{a}{b}x\). Calculate \(\frac{a}{b}\):\[\frac{\frac{1}{2}}{1} = \frac{1}{2}\]Thus, the asymptotes are \(y = \frac{1}{2}x\) and \(y = -\frac{1}{2}x\).

Step 5: Sketch the Graph

To sketch the hyperbola, draw the asymptotes as lines that cross the y-axis at the origin and have slopes of \(\frac{1}{2}\) and \(-\frac{1}{2}\). Mark the vertices and foci on the y-axis. Use the asymptotes as guides to draw the hyperbola opening upwards and downwards from the vertices.

Key Concepts

Understanding Hyperbola Vertices

In the realm of hyperbolas, vertices play a crucial role. They mark the points where the hyperbola intersects its axis of symmetry. For a vertical hyperbola, which revolves around the y-axis, these vertices lie on the y-axis.
Begin by identifying the standard form of a hyperbola equation:

• When in the form of \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the vertices are located at \((0, \pm a)\).

In our example, the solution found that \(a^2 = \frac{1}{4}\), which means \(a = \frac{1}{2}\). Therefore, the vertices are located at \((0, \frac{1}{2})\) and \((0, -\frac{1}{2})\). These points are the closest that the branches of the hyperbola come to one another on the y-axis.
Remember, vertices help sketch the hyperbola by pinpointing its narrowest points. These act as anchors, ensuring that the branches open symmetrically. Keep this in mind as you work with hyperbolas, as it's a consistent pattern across different forms.

Discovering the Foci of a Hyperbola

The foci of a hyperbola are another fundamental aspect, shaping the unique curves of the hyperbola's branches. For hyperbolas, these foci are found inside the arms of the curves, located farther from the center compared to the vertices.
The formula for the foci requires calculations of \(c\), which is found using the equation:

• \(c^2 = a^2 + b^2\)

In our equation, \(a^2 = \frac{1}{4}\) and \(b^2 = 1\), resulting in \(c^2 = \frac{1}{4} + 1 = \frac{5}{4}\). Solving for \(c\) gives \(c = \frac{\sqrt{5}}{2}\). So, the foci are located at \((0, \frac{\sqrt{5}}{2})\) and \((0, -\frac{\sqrt{5}}{2})\).
Understanding the position of the foci is vital for constructing the hyperbola. They lie along the y-axis and indicate where the hyperbola extends and contracts. While the vertices show us the narrowest points, the foci tell us how far the hyperbola stretches.

The Role of Asymptotes in Hyperbolas

Asymptotes in hyperbolas are invisible guideposts that help shape the graph's overall form. Each arm of the hyperbola approaches its corresponding asymptote but never touches it. Finding these lines provides a skeleton around which the hyperbola forms.
For vertical hyperbolas, the asymptote equations are:

• \(y = \pm \frac{a}{b}x\)

In this scenario, with \(a = \frac{1}{2}\) and \(b = 1\), the asymptotes are defined by the slopes \(y = \frac{1}{2}x\) and \(y = -\frac{1}{2}x\).
These lines cross the y-axis at the origin, serving as a crucial visual aid when sketching. Knowing the angles at which the hyperbola arms open—indicated by the slopes—ensures accurate depiction.
Asymptotes reinforce the symmetrical character of the hyperbola, ensuring balance and proper direction of its branches. When dealing with these components together, the hyperbola becomes much easier to understand and graph.