Problem 20 - 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. \(4 x^{2}-24 x-36 y^{2}-360 y+864=0\)

Step-by-Step Solution

Answer

The equation is \(\frac{(x-3)^2}{432} - \frac{(y+5)^2}{48} = 1\), vertices: \((3\pm 20.78, -5)\), foci: \((3\pm \sqrt{480}, -5)\), asymptotes: \(y + 5 = \pm \frac{1}{3}\sqrt{432}(x - 3)\)."

Step 1: Simplify the Equation

Start by rearranging and grouping the terms of the equation: \(4x^2 - 24x - 36y^2 - 360y + 864 = 0\). Move the constant term to the right side to simplify:\(4x^2 - 24x - 36y^2 - 360y = -864\).

Step 2: Complete the Square for x-terms

Factor out the coefficients of the squared term: \[4(x^2 - 6x) - 36(y^2 + 10y) = -864\]. First, complete the square for \(x^2 - 6x\): \((x^2 - 6x) = (x-3)^2 - 9\).

Step 3: Complete the Square for y-terms

Now, complete the square for \(y^2 + 10y\): \((y^2 + 10y) = (y+5)^2 - 25\). Substitute these completed squares back into the equation:\[4((x-3)^2 - 9) - 36((y+5)^2 - 25) = -864\].

Step 4: Simplify the Equation with Completed Squares

Simplify the equation after substituting the completed squares:\[4(x-3)^2 - 36 - 36(y+5)^2 + 900 = -864\].Combine constants:\[4(x-3)^2 - 36(y+5)^2 = -864 + 36 - 900\]\[4(x-3)^2 - 36(y+5)^2 = -1728\].

Step 5: Divide the Equation to Canonical Form

Divide every term by \(-1728\) to write the equation in standard form:\[\frac{(x-3)^2}{432} - \frac{(y+5)^2}{48} = 1\].This is the equation of a hyperbola with horizontal transverse axis.

Step 6: Identify the Vertices, Foci, and Asymptotes

For the hyperbola \(\frac{(x-3)^2}{432} - \frac{(y+5)^2}{48} = 1\), the center is \((3, -5)\), vertices are at \((3 + \sqrt{432}, -5)\) and \((3 - \sqrt{432}, -5)\). Calculate \(c\) using \(c^2 = a^2 + b^2\): \[c^2 = 432 + 48\], \[c = \sqrt{480}\].Asymptotes have the equation \(y + 5 = \pm \frac{b}{a}(x - 3)\), where \(b = 6.93\) and \(a = \sqrt{432} = 20.78\). Thus the asymptotes are:\(y + 5 = \pm \frac{1}{3}\sqrt{432}(x - 3)\).

Key Concepts

Understanding Hyperbolas

A hyperbola is a type of conic section formed when a plane intersects both nappes (or cones) of a double cone. They are often recognized for their distinctive U-shaped curves located on either side of the center point. Unlike parabolas or ellipses, hyperbolas consist of two separate branches. These branches open either horizontally or vertically, depending on the equation form. Hyperbolas have some unique properties:

• Two foci: Points within each branch where distances from any point on a branch are related.
• Vertices: Points on each branch where the hyperbola is closest to or farthest from the center.
• Asymptotes: Lines that the branches approach but never actually meet. These provide the hyperbola's directional "guide."

They are fundamental in studying different phenomena in physics and engineering, including waves, satellite orbits, and more.

Hyperbola in Standard Form

The standard form of a hyperbola's equation helps us quickly identify its properties and characteristics. For a hyperbola centered at \((h, k)\), the standard form can look like one of the following:

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

In these forms:

• \(a\) represents the distance from the center to each vertex along the major axis.
• \(b\) relates to the length from the center to the focal points on the minor axis.
• The relationship \(c^2 = a^2 + b^2\) helps determine the distance \(c\) from the center to the foci.

This form simplifies identifying key features like vertices, foci, and asymptotes. In the exercise, putting the equation in the standardized version revealed a hyperbola with a horizontal axis.

Vertices and Foci of Hyperbolas

The vertices and foci are crucial segments of a hyperbola, helping in its geometric description. For the hyperbola given in the exercise:

• Center: Located at \((3, -5)\).
• Vertices: Found at \((3 + \sqrt{432}, -5)\) and \((3 - \sqrt{432}, -5)\). The distance \(\sqrt{432}\) was derived by completing the square and simplifying the equation.
• Foci: The foci determine the opening width and are calculated using the relationship \(c^2 = a^2 + b^2\). For this hyperbola, \(c = \sqrt{480}\), and the foci are positioned at \((3 + \sqrt{480}, -5)\) and \((3 - \sqrt{480}, -5)\).

The foci lie along the major axis defined by the vertices. For horizontal hyperbolas, the foci will maintain a similar coordinate system, differing primarily in the best-known 'spread' factor \(c\).

Asymptotes of Hyperbolas

Asymptotes are vital lines that guide the shape and orientation of a hyperbola. They indicate the paths the hyperbola tends to, though they never intersect the curve itself. The asymptotes can be expressed as straight-line equations derived from the hyperbola's standard form. Based on the given hyperbola:

• The equation for asymptotes is \(y+5 = \pm \frac{b}{a}(x-3)\).
• Substituting known values from the problem where \(b = \sqrt{48} = 6.93\) and \(a = \sqrt{432} = 20.78\), the asymptotes become\( y + 5 = \pm \frac{6.93}{20.78}(x - 3)\).

This equation directly ties to the hyperbola's center at (3, -5), guiding both branches' trajectories. Asymptotes aid in sketching the curve and understanding its extent across the plane.