Problem 53 - Precalculus: Mathematics for Calculus

Question

Complete the square to determine whether the graph of the equation is an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an cllipse, find the center, foci, vertices, and lengths of the major and minor axes. If it is a parabola, find the vertex, focus, and directrix. If it is a hyperbola, find the center, foci, vertices, and asymptotes. Then sketch the graph of the equation. If the equation has no graph, explain why. $$16 x^{2}-9 y^{2}-96 x+288=0$$

Step-by-Step Solution

Answer

The equation represents a hyperbola with center (3, 0), vertices (0, 0) and (6, 0), and foci (-2, 0) and (8, 0).

Step 1: Rearrange and Group Terms

Start by moving the constant term to the other side of the equation. The given equation is \(16x^2 - 9y^2 - 96x + 288 = 0\). Subtract 288 from both sides to get \(16x^2 - 9y^2 - 96x = -288\). Now, group the \(x\) terms and \(y\) terms separately: \((16x^2 - 96x) - 9y^2 = -288\).

Step 2: Complete the Square for x-Terms

Factor out 16 from the \(x\) terms: \(16(x^2 - 6x)\). To complete the square, add and subtract \((\frac{-6}{2})^2 = 9\) inside the parenthesis: \(16(x^2 - 6x + 9 - 9)\) simplifies to \(16((x - 3)^2 - 9)\), which is \(16(x - 3)^2 - 144\).

Step 3: Combine and Rearrange the Equation

Now substitute the completed square back into the equation: \(16(x - 3)^2 - 144 - 9y^2 = -288\). Move constants to the other side: \(16(x - 3)^2 - 9y^2 = -288 + 144\), which simplifies to \(16(x - 3)^2 - 9y^2 = -144\).

Step 4: Divide by Coefficient of the Right Side

Divide the entire equation by \(-144\) to normalize it: \(\frac{16(x-3)^2}{-144} - \frac{9y^2}{-144} = 1\). Simplify this to \(-\frac{(x-3)^2}{9} + \frac{y^2}{16} = 1\).

Step 5: Identify Conic and Its Properties

The equation \(-\frac{(x-3)^2}{9} + \frac{y^2}{16} = 1\) is a hyperbola in the standard form \(-\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\). Here, \(h = 3\), \(k=0\), \(a^2 = 9\), and \(b^2 = 16\). The center is at \((3, 0)\). The vertices are at \((3\pm3, 0)\) or at the coordinates \((0, 0)\) and \((6, 0)\). The foci can be found using \(c^2 = a^2 + b^2\), hence \(c = \sqrt{25} = 5\), leading to foci at \((3\pm5, 0)\) or \((-2, 0)\) and \((8, 0)\). Asymptotes have slopes \(\pm \frac{b}{a} = \pm \frac{4}{3}\).

Step 6: Sketch the Graph

Sketch the hyperbola with the center \((3, 0)\), vertices at \((0, 0)\) and \((6, 0)\), and foci at \((-2, 0)\) and \((8, 0)\). Draw the asymptotes through the center with slopes \(\frac{4}{3}\) and \(-\frac{4}{3}\). The hyperbola will open left and right along the x-axis.

Key Concepts

Completing the Square

Completing the square is a mathematical technique used to simplify quadratic expressions and equations. It involves making a quadratic expression into a perfect square trinomial, which is a necessary step to identify the type of conic section an equation represents.
To complete the square for an expression like \(ax^2 + bx\), begin by factoring out the coefficient of \(x^2\) if it's not 1. Then, take the coefficient of \(x\), halve it, and square the result. This squared value is added and subtracted within the expression to form a perfect square trinomial. For example, if you have \(x^2 - 6x\):

• Halve \(-6\) to get \(-3\), and then square it to get \(9\).
• Add \(9\) and simultaneously subtract \(9\) to maintain equation balance: \(x^2 - 6x + 9 - 9\).
• Re-organize it to form \((x - 3)^2 - 9\).

This method transforms the expression into a form that is easier to work with when identifying conic sections in an equation.

Hyperbola

A hyperbola is one of the four types of conic sections, which also include circles, ellipses, and parabolas. It is a curve formed by the intersection of a double cone and a plane where the angle of the plane is less steep than that of the cone. The standard equation for a hyperbola is:
\[ -\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]
Here, \((h, k)\) represents the center of the hyperbola. A hyperbola has two branches that mirror each other about a central axis. Unlike ellipses, hyperbolas open outward either horizontally or vertically. In this context, the hyperbola's symmetry and shape are determined by the terms \(a^2\) and \(b^2\).

• If a hyperbola opens horizontally, the equation form is \(-\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\).
• If it opens vertically, the equation becomes \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\).
• In our given problem, it opens horizontally since the \(x\)-term is negative.

The graph of a hyperbola consists of two separate curves that move farther apart as they extend.

Vertices and Foci of Conics

In the realm of conic sections, vertices and foci are critical to understanding the shape and orientation of these curves. Vertices are points where each branch of the conic has its sharpest turn. For hyperbolas, they are located on the axis that connects the two branches.

• For the hyperbola formed by the equation \( -\frac{(x-3)^2}{9} + \frac{y^2}{16} = 1 \), the vertices are found at \((3 \pm 3, 0)\) which results in points \((0, 0)\) and \((6, 0)\).
• The formula used here is \(\text{Vertices} = (h \pm a, k)\).
• The foci of a hyperbola are points that lie along the axis with greater distance from the center than the vertices, found using \(c^2 = a^2 + b^2\), giving \(c = 5\) in this scenario, leading to foci at \((3 \pm 5, 0)\), or \((-2, 0)\) and \((8, 0)\).
• The foci are essential in defining the hyperbola's shape since each point on the curve produces constant difference of distances to them.

Asymptotes can also be calculated for hyperbolas, which guide their opening direction.

Standard Form of a Conic Equation

The standard form of a conic section equation allows you to easily identify the type of conic section represented. Conics can be expressed in several standard forms, and identifying the correct one helps in graphing and solving related problems.
Hyperbolas, for example, have a standard form of
\[ -\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]
This standard equation indicates that we are dealing with a hyperbola.

• For ellipses, the equation would be \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), resembling a hyperbola but instead involves addition.
• Inthe case of parabolas, the general form becomes \((x-h)^2 = 4p(y-k)\) or \((y-k)^2 = 4p(x-h)\), depending on orientation.

Transforming and arranging equations into their standard form is crucial for understanding each conic section's characteristics, as it allows one to determine key geometric properties like vertices, foci, and orientation at a glance.