Problem 31 - 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. $$x^{2}-9 y^{2}+2 x-54 y-80=0$$

Step-by-Step Solution

Answer

The center is at (1,-3), the vertices are at (1±\(\sqrt{99}\), -3), the foci are at (1±\(\sqrt{110}\), -3), and the equations of the asymptotes are \(y = -3 ± \frac{\sqrt{11}}{\sqrt{99}}(x - 1)\). The hyperbola opens to the left and right.

Step 1: Put the given equation in the standard form

Complete the square to put the equation into standard form of a hyperbola, which is \(\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\), where (h, k) is the center of the hyperbola. Rearranging the terms and completing squares, we get \((x-1)^2-9(y+3)^2=99\). Divide all terms by 99 to make the right side equal to 1. Hence, we get \(\frac{(x-1)^2}{99}-\frac{(y+3)^2}{11}=1\)

Step 2: Identify the center, vertices, and foci

From standard form, center (h, k) is at (1, -3). Vertices are distance 'a' from the center, which is \(\sqrt{99}\). Therefore, the vertices are at (1±\(\sqrt{99}\), -3). The foci are at a distance 'c' from the center where \(c=\sqrt{a^2+b^2} = \sqrt{99+11} = \sqrt{110}\). Thus, the foci are at (1±\(\sqrt{110}\), -3).

Step 3: Find asymptotes

The equations of the asymptotes are given by \(y = k ± \frac{b}{a}(x - h)\). Substituting the values of a, b, h, and k, we have \(y = -3 ± \frac{\sqrt{11}}{\sqrt{99}}(x - 1)\).

Step 4: Sketch the hyperbola

Plot the center at point (1,-3). Sketch the two vertices and foci. Draw the rectangle that the hyperbola passes through at the vertices and whose diagonals are the asymptotes. Use asymptotes as guideline to sketch branches of the hyperbola. You will notice the graph of hyperbola opening left and right.

Key Concepts

Asymptotes

Asymptotes are like invisible guides that help us to sketch a hyperbola more accurately. They are straight lines that a hyperbola approaches but never actually touches. Asymptotes act as a boundary displaying the direction in which the hyperbola opens as it stretches infinitely outward.
To find the equations of the asymptotes for a hyperbola given in the form \(\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\), we use the formula:

• \(y = k \pm \frac{b}{a}(x - h)\)

Plugging the values from the standard form, we derived the asymptotes’ equations as \(y = -3 \pm \frac{\sqrt{11}}{\sqrt{99}}(x - 1)\). These lines will never meet the curves of the hyperbola but help us draw its full shape.

Center of Hyperbola

The center of a hyperbola is the point from which its open arms spread out. It’s the pivotal point that determines the positioning of the hyperbola's vertices and foci.
In a hyperbola equation like:

• \(\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\)

The center is represented as \((h,k)\).
From the equation \(\frac{(x-1)^2}{99}-\frac{(y+3)^2}{11}=1\), the center is calculated to be \((1, -3)\). This center acts as the anchor of the hyperbola, aligning the location of vertices and foci symmetrically around it.

Vertices of Hyperbola

The vertices of a hyperbola are the closest points of each branch to its center. They define the width of the hyperbola and help us determine its orientation, whether it opens horizontally or vertically.
In a standard form equation:

• \(\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\)

The vertices are located a distance \(a\) away from the center \((h,k)\). For this hyperbola, \(a = \sqrt{99}\), so the vertices lie at points \((1 \pm \sqrt{99}, -3)\).
These vertices provide key reference points when sketching the hyperbola.

Foci of Hyperbola

Foci are special points that, despite not lying on the hyperbola, play a crucial role in forming its shape. They sit inside each branch of the hyperbola and help to define the curves’ steepness and width.
To locate the foci given a hyperbola in the form:

• \(\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\)

We calculate the distance \(c\) from the center using \(c = \sqrt{a^2 + b^2}\). For this particular hyperbola, \(c = \sqrt{110}\). Thus, the foci are placed at \((1 \pm \sqrt{110}, -3)\).
Having these coordinates allows us to better understand the hyperbola’s overall structure.

Standard Form of Hyperbola

The standard form of a hyperbola helps us solve and graph hyperbolas efficiently, giving us essential insights like the center, vertices, and direction. This standard form is:

• \(\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\)

Originally, the equation \(x^2-9y^2+2x-54y-80=0\) was not in standard form. By rearranging and completing the square, we turned it into \(\frac{(x-1)^2}{99}-\frac{(y+3)^2}{11}=1\).
The equation demonstrates that the hyperbola is horizontally oriented since \(a^2 > b^2\). This simple format allows us to extract meaningful data about its graph and guide its precise sketching.