Problem 28 - Precalculus

Question

Find the center, vertices, foci, and the equations of the asymptotes of the hyperbola. Then sketch the hyperbola using the asymptotes as an aid. $$\frac{(y-1)^{2}}{1 / 4}-\frac{(x+3)^{2}}{1 / 16}=1$$

Step-by-Step Solution

Answer

The center of the hyperbola is \(-3, 1\). The vertices are \(-3, 3/2\) and \(-3, 1/2\). The foci are \(-3, 21/16\) and \(-3, 11/16\). The equations of the asymptotes are \(y = 2x + 7\) and \(y = -2x - 5\).

Step 1: Identify the Center of the Hyperbola

First, identify the center of the hyperbola from the given equation. It is the set \((h, k)\), which is \((-3, 1)\) here.

Step 2: Find the Vertices

Next, find the vertices, which can be found at \((h, k \pm a)\). Here, \(a^{2}=1/4\) or \(a=1/2\), which gives vertices as \((-3, 1+1/2)\) and \((-3, 1-1/2)\) or \((-3, 3/2)\) and \((-3, 1/2)\).

Step 3: Find the Foci

The foci are located at \((h, k \pm c)\), where \(c^{2} = a^{2} + b^{2}\). Here, \(b^{2} = 1/16\) or \(b=1/4\), so \(c = \sqrt{a^2+b^2} = \sqrt{1/4 + 1/16} = \sqrt{4/16 + 1/16} = \sqrt{5/16} = 5/16\). Thus, foci are at \((-3, 1+ 5/16)\) and \((-3, 1- 5/16)\) or \((-3, 21/16)\) and \((-3, 11/16)\).

Step 4: Find the Equations of the Asymptotes

The equations of the asymptotes can be found using the formula \(y = k \pm \frac{a}{b}(x - h)\). Thus, the asymptotes are \(y = 1 \pm \frac{1/2}{1/4}(x + 3)\) or \(y = 1 \pm 2(x + 3)\), resulting in two equations, \(y = 2x + 7\) and \(y = -2x - 5\).

Step 5: Sketch the Hyperbola

Lastly, sketch the hyperbola by first plotting the center. Then plot the vertices and asymptotes, and draw the hyperbola opening upwards and downwards about the center. The foci will be inside the hyperbola.

Key Concepts

Asymptotes of a Hyperbola

Asymptotes are straight lines that a hyperbola approaches but never touches. They provide a framework to help us sketch the hyperbola accurately. To find the equations of the asymptotes for the given hyperbola \(\frac{(y-1)^{2}}{1 / 4}-\frac{(x+3)^{2}}{1 / 16}=1\), we use the formula \(y = k \pm \frac{a}{b}(x - h)\).

For our hyperbola, \(a = \frac{1}{2}\) and \(b = \frac{1}{4}\), making the center at \((-3, 1)\). Plugging these values into our formula yields two equations, which become our asymptotes: \(y = 1 \pm 2(x + 3)\). Simplifying these, we get \(y = 2x + 7\) and \(y = -2x - 5\).

Understanding that these lines provide the 'boundary' behavior for the hyperbola is crucial. They will not intersect the hyperbola at any point but serve as a guide to comprehend where the hyperbola will be located and how it will 'open'. In the sketching process, they help us visualize the extent of the hyperbola.

Vertices of a Hyperbola

Vertices are essential points that determine the most prominent features of a hyperbola. For the exercise hyperbola \(\frac{(y-1)^{2}}{1 / 4}-\frac{(x+3)^{2}}{1 / 16}=1\), the vertices are found using the center location \((h, k)\) and the value of \(a\).

Specifically, the vertices are located at \((h, k \pm a)\). Considering our center \((-3, 1)\) and with \(a = \frac{1}{2}\), we find our vertices at \((-3, 1 + \frac{1}{2})\) and \((-3, 1 - \frac{1}{2})\), which simplifies to \((-3, \frac{3}{2})\) and \((-3, \frac{1}{2})\).

The vertices mark the closest points of the hyperbola to the center—they lie along the transverse axis, which is the longest line segment that can be drawn through the center of the hyperbola. These points are crucial when sketching the curve as they help to determine the shape and size of the hyperbola.

Foci of a Hyperbola

The foci (singular: focus) of a hyperbola are two points that are used to define the curve mathematically. For our given hyperbola equation, we can find the foci by identifying \(c\), which is the distance from each focus to the center, using the formula \(c^{2} = a^{2} + b^{2}\).

For our hyperbola with \(a^{2} = \frac{1}{4}\) and \(b^{2} = \frac{1}{16}\), we have \(c = \sqrt{\frac{1}{4} + \frac{1}{16}} = \sqrt{\frac{5}{16}} = \frac{5}{4}\). This tells us the foci are located at \((h, k \pm c)\), specifically at \((-3, 1 + \frac{5}{4})\) and \((-3, 1 - \frac{5}{4})\), or \((-3, \frac{21}{16})\) and \((-3, \frac{11}{16})\).

The foci are crucial for understanding the properties of a hyperbola because the difference in distances from any point on the hyperbola to the foci is constant. This unique characteristic illustrates how the curve is shaped around these points, and they are considered when sketching and analyzing the behavior of the hyperbola.