Problem 21 - College Algebra with Corequisite Support

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. \(-4 x^{2}+24 x+16 y^{2}-128 y+156=0\)

Step-by-Step Solution

Answer

The standard form is \(\frac{(y-4)^2}{4} - \frac{(x-3)^2}{16} = 1\). Vertices: (3, 2) and (3, 6). Foci: (3, 1.6) and (3, 6.4). Asymptotes: \(y = \frac{1}{2}x + 2.5\) and \(y = -\frac{1}{2}x + 5.5\).

Step 1: Rearrange and Group Terms

Start by rearranging the terms in the equation to group the x-terms together and the y-terms together: \[-4x^2 + 24x + 16y^2 - 128y = -156\].

Step 2: Factor out the Coefficients

Factor out the coefficients of the squared terms to prepare for completing the square:\[ -4(x^2 - 6x) + 16(y^2 - 8y) = -156 \].

Step 3: Complete the Square for x-terms

For the x-terms inside the parentheses, complete the square:\[ x^2 - 6x \] becomes \[ (x - 3)^2 - 9 \]. Hence, the equation becomes:\[ -4((x - 3)^2 - 9) + 16(y^2 - 8y) = -156 \].

Step 4: Complete the Square for y-terms

For the y-terms inside the parentheses, complete the square:\[ y^2 - 8y \] becomes \[ (y - 4)^2 - 16 \]. So, the equation now is:\[ -4((x - 3)^2 - 9) + 16((y - 4)^2 - 16) = -156 \].

Step 5: Simplify the Equation

Distribute and simplify the equation:\[-4(x - 3)^2 + 36 + 16(y - 4)^2 - 256 = -156 \].Combine constants to get:\[ -4(x - 3)^2 + 16(y - 4)^2 = 64 \].

Step 6: Write in Standard Form

Divide the entire equation by 64 to set the equation to standard form:\[ \frac{(y-4)^2}{4} - \frac{(x-3)^2}{16} = 1 \].Now, this is the standard form of a hyperbola with vertical transverse axis.

Step 7: Identify the Vertices and Foci

For the hyperbola \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \), the center is \((h, k)\).Here, the center is \((3, 4)\), vertices are \((3, 4 \pm 2) = (3, 2)\) and \((3, 6)\), and foci are \((3, 4 \pm c)\) where \(c = \sqrt{a^2 + b^2} = \sqrt{4 + 16} = \sqrt{20}\), giving foci at approximately \((3, 1.6)\) and \((3, 6.4)\).

Step 8: Find the Asymptotes

The asymptotes for this hyperbola are given by the slopes \(\pm \frac{a}{b}\).Here, \(a = 2\) and \(b = 4\). So the equation for asymptotes are:\[ y - 4 = \pm \frac{2}{4}(x - 3) \].Simplified, the equations become:\[ y = \frac{1}{2}x + 2.5 \] and \[ y = -\frac{1}{2}x + 5.5 \].

Key Concepts

Vertices of Hyperbola

In the study of hyperbolas, vertices play a crucial role. Imagine a hyperbola as two symmetrical curves. The vertices are essentially the points where these curves come closest together. They sit, in simpler terms, at the dip in the curve.For hyperbolas centered at \( (h, k) \), various formulas help find these vertices based on their orientation:

• For a vertical hyperbola, as we have here, the vertices positions are \( (h, k \pm a) \).
• For a horizontal hyperbola, they'd be \( (h \pm a, k) \).

Here, our hyperbola is vertically oriented with a center at \( (3, 4) \), and \( a = 2 \) as derived from \( \frac{(y-4)^2}{4} \). Hence, our vertices become \( (3, 4 \pm 2) \). This means their exact positions are \( (3, 2) \) and \( (3, 6) \). These points can intuitively help model the extent and shape of the hyperbola on a graph.

Foci of Hyperbola

The foci are like the guiding stars of the hyperbola, influencing its overall shape and orientation. Just as ellipse foci determine the pull of the curve inward, hyperbola foci dictate how outward and extended the arms are.In the equation provided, where the center is \( (3, 4) \), the calculation of foci involves another segment of the hyperbola's attributes. Here, \( a = 2 \) and \( b = 4 \), leading to calculating \( c \) using the formula \( c = \sqrt{a^2 + b^2} \). So, \( c = \sqrt{4 + 16} = \sqrt{20} \).The foci for vertical hyperbolas follow the positions \( (h, k \pm c) \). Applying our values, the foci lay approximately at \( (3, 4 \pm \sqrt{20}) \), which evaluates roughly to \( (3, 1.6) \) and \( (3, 6.4) \). Unlike vertices, foci are not typically marked explicitly on a graph, but understanding their positions augments comprehension of hyperbola layout.

Asymptotes of Hyperbola

Asymptotes guide the potential path of a hyperbola's arms, signifying directions toward which the tails extend, never officially touching these lines. In essence, asymptotes help describe the expanding nature of hyperbolic arms.The calculation for asymptotes involves both \( a \) and \( b \) values. Using the slope \( \pm \frac{a}{b} \), where \( a = 2 \) and \( b = 4 \), the slopes become \( \pm \frac{1}{2} \).Hence, for our vertical hyperbola, the asymptote equations transpose into:

• \( y - 4 = \frac{1}{2}(x - 3) \)
• \( y - 4 = -\frac{1}{2}(x - 3) \)

Simplifying these gives:

• \( y = \frac{1}{2}x + 2.5 \)
• \( y = -\frac{1}{2}x + 5.5 \)

These equations elucidate the slant of the hyperbola arms on either side, providing a directional compass of sorts within the context of a coordinate plane.