Problem 38 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the standard form of the equation of the hyperbola, (b) find the center, vertices, foci, and asymptotes of the hyperbola, and (c) sketch the hyperbola. Use a graphing utility to verify your graph. \(3 y^{2}-5 x^{2}+6 y-60 x-192=0\)

Step-by-Step Solution

Answer

Standard form: \(\frac{(y+1)^2}{99}-\frac{(x+6)^2}{59}=1 \). \nCenter: (-6, -1), Vertices: (-6,-1±\(\sqrt{99}\)), Foci: (-6,-1± \(\sqrt{158}\)). Asymptotes: \( \pm \frac{(y+1)}{(x+6)}= \sqrt{\frac{99}{59}} \)

Step 1: Rewrite the equation in standard form

Rearrange the equation and group the terms: \(3(y^2 + 2y)-5(x^2 + 12x) = 192. \) \nNow, complete the square for each parantheses: \nIt becomes \(3[(y+1)^2 -1] - 5[(x+6)^2 - 36] = 192 \n=> 3(y+1)^2 -5(x+6)^2 = 297. \) \nFinally, to get the standard form divide every term by 297 so the right end becomes 1:\n\(\frac{(y+1)^2}{99}-\frac{(x+6)^2}{59} =1. \)

Step 2: Identify the features of the hyperbola

Center: The center is given by the equations \(y=-1, x=-6 \), so it's at (-6, -1)\nVertices: The vertices are on the y-axis, a distance of \(\sqrt{99} \) away from the center. So vertices are at (-6,-1±\(\sqrt{99}\)) \nFoci: The foci are also on the y-axis and given by \(\pm \sqrt{99+59} \) away from the center. So foci are at (-6,-1± \(\sqrt{158}\)). \nAsymptotes: Given by the equation \( \pm \frac{(y+1)}{(x+6)}= \sqrt{\frac{99}{59}}. \)

Step 3: Draw the hyperbola

First draw the center, vertices and foci. Then draw the asymptote lines according to their equation. After that, draw a smooth curve that passes through the vertices and the foci, indicating the shape of the hyperbola. To verify the graph, make use of a graphing utility.

Key Concepts

Standard Form of Hyperbola

A hyperbola has a standard form that makes it easier to identify its characteristics. It is similar to other conic sections, like ellipses, because it involves squared terms. However, hyperbolas differ because one term is subtracted in the equation. The general standard form of a hyperbola is:

• For a vertical transverse axis: \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \)
• For a horizontal transverse axis: \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \)

Here,

• \((h, k)\) is the center of the hyperbola.
• \(a\) and \(b\) are distances related to the size and orientation of the hyperbola.

To get the equation into standard form, you'll need to rearrange and complete the square. In the provided problem, the standard form becomes: \[ \frac{(y+1)^2}{99} - \frac{(x+6)^2}{59} = 1 \]

Center of Hyperbola

The center of a hyperbola is a fundamental feature, serving as the anchor for other components like vertices and asymptotes. In the standard form equations, the center is located at \((h, k)\). It moves from the origin to this point, influencing the entire graph's position. From the original equation:

• The center is found at \((-6, -1)\).

Knowing the center is crucial because it helps plot the hyperbola on a coordinate plane precisely, aiding in sketching and analysis.

Vertices and Foci of Hyperbola

Vertices and foci are key elements that define the hyperbola's shape. They lie along the transverse axis, either horizontally or vertically, depending on the form of the equation.

Vertices

• The vertices are \(a\) units from the center along the transverse axis. In this exercise, they are located at: \((-6, -1 \pm \sqrt{99})\).

Foci

• The foci are additional points that are \(c\) units from the center, where \(c^2 = a^2 + b^2\).
• Using the given equation: \(c = \sqrt{158} \).Thus, the foci are at:\((-6, -1 \pm \sqrt{158})\).

The placement of vertices and foci dictates the hyperbola's branches and width.

Asymptotes of Hyperbola

Asymptotes are lines that the hyperbola approaches but never touches, helping to visualize its open ends. They form an 'X' through the center and are essential in sketching the hyperbola accurately. The asymptotes follow the equation: \[ \frac{(y-k)}{a} = \pm \frac{(x-h)}{b} \]Given the transformation from the problem:

• The asymptotes are determined by: \( \pm \frac{(y+1)}{(x+6)} = \sqrt{\frac{99}{59}} \).

Using this formula, you can understand how quickly the hyperbola spreads outwards. The slopes from the equation reveal the angle and steepness of the asymptotes, guiding the overall sketch.