Problem 30 - College Algebra

Question

Find the equations of the asymptotes for each hyperbola. \(16 y^{2}+96 y-4 x^{2}+16 x+112=0\)

Step-by-Step Solution

Answer

The equations of the asymptotes are \(y = \frac{1}{2}x - 4\) and \(y = -\frac{1}{2}x - 2\).

Step 1: Rewrite the Equation in Standard Form

First, we begin by rewriting the given equation of the hyperbola in its standard form. The given equation is:\[16y^{2}+96y-4x^{2}+16x+112=0\]Group the terms involving \(y\) and \(x\):\[16(y^{2}+6y) - 4(x^{2}-4x) + 112 = 0\]

Step 2: Complete the Square

Complete the square for both the \(y\) and \(x\) terms:For the \(y\) terms: \[y^{2}+6y \Rightarrow (y+3)^{2} - 9\]For the \(x\) terms:\[x^{2}-4x \Rightarrow (x-2)^{2} - 4\]Substitute these squared forms back into the equation:\[16((y+3)^{2} - 9) - 4((x-2)^{2} - 4) + 112 = 0\]Simplify:\[16(y+3)^{2} - 144 - 4(x-2)^{2} + 16 + 112 = 0\]This reduces to:\[16(y+3)^{2} - 4(x-2)^{2} - 16 = 0\]

Step 3: Express in Standard Hyperbola Form

Rearrange the equation to isolate the zero and factor out the constants:\[16(y+3)^{2} - 4(x-2)^{2} = 16\]Divide the entire equation by 16 to simplify:\[\frac{(y+3)^{2}}{1} - \frac{(x-2)^{2}}{4} = 1\]Our hyperbola is now in its standard form, \(\frac{(y-k)^{2}}{a^2} - \frac{(x-h)^{2}}{b^2} = 1\), where \(a^2 = 1\) and \(b^2 = 4\).

Step 4: Find the Asymptotes

To find the equations of the asymptotes for a hyperbola in the form \(\frac{(y-k)^{2}}{a^2} - \frac{(x-h)^{2}}{b^2} = 1\), we use:\[y - k = \pm \frac{a}{b}(x - h)\]Substituting \(k = -3\), \(a = 1\), \(b = 2\), and \(h = 2\) into the asymptote formula, we get:\[y + 3 = \pm \frac{1}{2}(x - 2)\]These simplify to the equations:\[y + 3 = \frac{1}{2}(x - 2)\]And\[y + 3 = -\frac{1}{2}(x - 2)\]

Step 5: Simplify the Asymptotes

Simplifying these equations, \[y + 3 = \frac{1}{2}x - 1\]Leads to:\[y = \frac{1}{2}x - 4\]And for the second equation:\[y + 3 = -\frac{1}{2}x + 1\]Which simplifies to:\[y = -\frac{1}{2}x - 2\]

Key Concepts

Understanding Hyperbola

A hyperbola is a set of all points in the plane where the difference of the distances from two fixed points (the foci) is constant. It looks like two mirrored curves or an elongated X shape. These curves are in opposition to each other and open either horizontally or vertically. Unlike parabolas or circles, hyperbolas have two disjoint curves called branches. The hyperbola's center, not located at either of the two branches, is crucial in defining its standard form.

The hyperbola's orientation depends on the signs in its equation. When the \(rac{(y-k)^2}{a^2} - rac{(x-h)^2}{b^2} = 1\), it opens vertically; while \(rac{(x-h)^2}{a^2} - rac{(y-k)^2}{b^2} = 1\) opens horizontally. The terms inside the fractions determine how the hyperbola scales across the coordinate grid.

Completing the Square

Completing the square is a technique used to make an equation easier to work with by converting it into a form that reveals essential features like its center. This mathematical trick is used primarily for quadratic equations, hyperbolas, and circles.

Here's a simple way to understand completing the square:

• Identify the quadratic equation you need to transform.
• Group the quadratic and linear terms.
• Take half the coefficient of the linear term, square it, and add or subtract it within the group.

In this exercise, for the \(y\)-terms a square is completed as follows: \(y^2 + 6y \Rightarrow (y+3)^2 - 9\), and similarly for the \(x\)-terms: \(x^2 - 4x \Rightarrow (x-2)^2 - 4\).

This restructuring is essential to put the equation into a recognizable standard form, which helps in further analysis.

Standard Form of Hyperbola

The standard form is vital because it simplifies identifying properties like the center, vertices, and asymptotes. For hyperbolas, the standard form is generally expressed as \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\) or \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\). Here, \(h\) and \(k\) denote the center of the hyperbola.

In our example, after completing the square, we obtained:\(\frac{(y+3)^2}{1} - \frac{(x-2)^2}{4} = 1\). This tells us:

• The center of the hyperbola is at \((h, k) = (2, -3)\)
• The square root of the numerator in each term provides the transverse axis' length, and the conjugate axis is represented by the denominators \(a\) and \(b\).

This form helps derive the equation of asymptotes and locate other critical features.

Equations of Asymptotes

Asymptotes are invisible boundary lines guiding the branches of the hyperbola's opening. For hyperbolas in the form \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\), the asymptotes are expressed as:\(y - k = \pm \frac{a}{b}(x - h)\).

These lines extend through the center and provide guidelines, suggesting how quickly each branch of a hyperbola diverges. In practice, these equations contain the same slope but differ in direction (positive and negative).

• For our instance, substituting \(k = -3\), \(a = 1\), \(b = 2\), and \(h = 2\) results in \(y + 3 = \pm \frac{1}{2}(x - 2)\).

Simplifying gives precise expressions of the asymptotes:\(y = \frac{1}{2}x - 4\) and \(y = -\frac{1}{2}x - 2\). These lines are fundamental in graphing the hyperbola accurately, offering a graphical sense that aligns with mentioned algebraic descriptions.