Problem 20 - College Algebra

Question

For the following exercises, write the equation for the hyperbola in standard form if it is not already, and identify the vertices and foci, and write equations of asymptotes. $$ 4 x^{2}-24 x-36 y^{2}-360 y+864=0 $$

Step-by-Step Solution

Answer

Equation: \((x-3)^2/9 - (y+5)^2/4 = 1\); Vertices: (0,-5), (6,-5); Foci: \((3\pm\sqrt{13},-5)\); Asymptotes: \(y=-5\pm\frac{2}{3}(x-3)\).

Step 1: Reorganize the equation

First, let's reorganize the terms of the equation to separate the variables and constants. Start with:\[4x^2 - 24x - 36y^2 - 360y + 864 = 0\]Rearrange terms: \[4x^2 - 24x - 36y^2 - 360y = -864\]

Step 2: Factor and complete the square

Factor out the coefficients of the quadratic terms:\[4(x^2 - 6x) - 36(y^2 + 10y) = -864\]Complete the square:For x: \[x^2 - 6x\] becomes \[(x-3)^2 - 9\], so add 9 inside the parentheses and outside multiply by 4.For y: \[y^2 + 10y\] becomes \[(y+5)^2 - 25\], so add 25 inside the parentheses and outside multiply by 36.

Step 3: Rewrite the completed square equation

Substitute completed squares back:\[4((x-3)^2 - 9) - 36((y+5)^2 - 25) = -864\]Distribute and simplify:\[4(x-3)^2 - 36(y+5)^2 - 36 + 900 = -864\]Combine like terms and simplify:\[4(x-3)^2 - 36(y+5)^2 = 0\]Which simplifies further to:\[(x-3)^2 - 9(y+5)^2 = 0\]

Step 4: Write in standard form of hyperbola

Divide whole equation by coefficient:\[\frac{(x-3)^2}{9} - \frac{(y+5)^2}{4} = 1\]Now, the equation is in standard form: \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\]where \((h, k) = (3, -5)\), \(a = 3\), \(b = 2\).

Step 5: Identify vertices and foci

Use \((h, k) = (3, -5)\). The horizontal transverse axis means vertices are at \((h\pm a, k)\) => \((3 \pm 3, -5)\) = (0, -5) and (6, -5).For foci, use equation: \(c = \sqrt{a^2 + b^2}\) => \(c = \sqrt{9 + 4} = \sqrt{13}\).So, foci are \((h\pm c, k)\) => \((3 \pm \sqrt{13}, -5)\).

Step 6: Write equations for the asymptotes

The equations of the asymptotes for the horizontal hyperbola \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) are:\[y = k \pm \frac{b}{a}(x-h)\], thus:\[y = -5 \pm \frac{2}{3}(x-3)\].

Key Concepts

Standard Form

In the study of hyperbolas, the standard form of their equation is crucial for understanding their geometry. A hyperbola can be represented in the standard form:

• If it is horizontal: \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \)
• If it is vertical: \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \)

Here, \((h, k)\) represents the center of the hyperbola, while \(a\) and \(b\) are distances related to the shape and spread of the hyperbola. In the problem, after completing the square and organizing terms, we derived the standard form:\[ \frac{(x-3)^2}{9} - \frac{(y+5)^2}{4} = 1 \]By achieving this form, it is easy to determine other important characteristics of the hyperbola, like vertices, foci, and asymptotes.

Vertices

In a hyperbola, vertices are key points that lie on either side of the center along the axes. For the standard form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), the vertices are determined based on the orientation:

• For a horizontal hyperbola: vertices are at \((h \pm a, k)\)
• For a vertical hyperbola: vertices are at \((h, k \pm a)\)

Given this exercise's hyperbola, we found vertices from the achieved standard form equation. Since it is horizontal, the vertices are calculated as:\[ (h\pm a, k) = (3 \pm 3, -5) \]This results in vertices at points (0, -5) and (6, -5). These points indicate the "ends" of the hyperbola along the designated axis.

Foci

The foci of a hyperbola are points outside the curve, from which the "stretch" of the hyperbola is defined. For a hyperbola described by the standard form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), the distance to the foci from the center is given by \( c = \sqrt{a^2 + b^2} \). The foci are located differently based on the orientation:

• For a horizontal hyperbola: at \((h \pm c, k)\)
• For a vertical hyperbola: at \((h, k \pm c)\)

Using the exercise's standard form, and knowing \( a = 3 \) and \( b = 2 \), we calculate:\[ c = \sqrt{9 + 4} = \sqrt{13} \]Therefore, foci positions are:\[ (h \pm c, k) = (3 \pm \sqrt{13}, -5) \]These positions help to understand the symmetry and drawing of the hyperbola.

Asymptotes

Asymptotes in a hyperbola serve as guide lines that the branches of the hyperbola approach but never touch. They define the hyperbola's openness and direction. In the standard form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), the equations for asymptotes depend on the orientation:

• For a horizontal hyperbola: \( y = k \pm \frac{b}{a}(x-h) \)
• For a vertical hyperbola: \( y = k \pm \frac{a}{b}(x-h) \)

For the hyperbola in our problem, using the determined values \( a = 3 \) and \( b = 2 \), the asymptotes are:\[ y = -5 \pm \frac{2}{3}(x-3) \]These lines, \( y = -5 + \frac{2}{3}(x-3) \) and \( y = -5 - \frac{2}{3}(x-3) \), serve as boundary lines that the hyperbola approaches. Understanding these equations helps in graphing the hyperbola and comprehending its spatial behavior.