Problem 29 - Precalculus Mathematics for Calculus

Question

Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellipse, 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), foci (-2, 0) and (8, 0), and asymptotes \(y=\pm\frac{4}{3}(x-3)\).

Step 1: Rearrange the Equation

Start by rearranging the terms of the equation to focus on like-variable groups.Move all the terms to one side of the equation:\[16x^2 - 96x - 9y^2 + 288 = 0\]

Step 2: Isolate the Constant Term

Move the constant term to the right side of the equation:\[16x^2 - 96x - 9y^2 = -288\]

Step 3: Group Like Terms

Group the terms involving the same variable:\[16(x^2 - 6x) - 9y^2 = -288\]

Step 4: Complete the Square for x

Complete the square for the terms involving \(x\):* For the \(x\)-terms: Take \(-6\), divide by 2 to get \(-3\), and square it to get 9. Add and subtract this in the expression:\[16(x^2 - 6x + 9 - 9) - 9y^2 = -288\]Rearrange to complete the square:\[16((x - 3)^2 - 9) - 9y^2 = -288\]

Step 5: Simplify the Expression

Distribute and simplify:\[16(x - 3)^2 - 144 - 9y^2 = -288\]Add 144 to both sides:\[16(x - 3)^2 - 9y^2 = -144\]Divide the entire equation by -144 to simplify:\[\frac{16(x - 3)^2}{-144} - \frac{9y^2}{-144} = 1\]

Step 6: Identify the Equation Type

Simplify the expression:\[\frac{(x - 3)^2}{9} - \frac{y^2}{16} = 1\]This is in the standard form of a hyperbola \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) with the hyperbola opening in the horizontal direction.

Step 7: Find the Hyperbola's Center, Foci, and Vertices

The center \((h, k)\) is \((3, 0)\).The vertices are found along the x-axis, \((h-a, k)\) and \((h+a, k)\), i.e., \((0, 0)\) and \((6, 0)\).Foci are found using \(c^2 = a^2 + b^2\), so \(c = \sqrt{9 + 16} = 5\) giving foci at \((h-c, k)\) and \((h+c, k)\), i.e., \((-2, 0)\) and \((8, 0)\).

Step 8: Find Asymptotes of the Hyperbola

The asymptotes are given by the slopes \(\frac{b}{a}\) and \(-\frac{b}{a}\). Here, \(a = 3\) and \(b = 4\), thus the slopes of the asymptotes are \(\frac{4}{3}\) and \(-\frac{4}{3}\). The asymptote equations are:\[y = \frac{4}{3}(x-3)\] and \[y = -\frac{4}{3} (x-3)\]

Step 9: Sketch the Hyperbola

Draw the center at \((3, 0)\), mark the vertices at \((0, 0)\) and \((6, 0)\), plot the foci at \((-2, 0)\) and \((8, 0)\), and draw the asymptotes through the center depicting the 'X' shape. The hyperbola opens horizontally.

Key Concepts

Hyperbola

The hyperbola is one of the four types of conic sections—shapes formed by the intersection of a plane and a double-napped cone. The standard form of a hyperbola that opens horizontally is given by the equation:

• \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\]

Here,

• \((h, k)\) is the center of the hyperbola,
• \(a\) is the distance from the center to each vertex along the line of symmetry,
• \(b\) is the distance related to the slope of the asymptotes.

To distinguish a hyperbola from other conic sections, notice the minus sign between the squares of \(x\) and \(y\). In a hyperbola that opens horizontally, the \(x\)-terms have a positive fraction, while the \(y\)-terms have a negative fraction. This shape creates two branches that look like mirror images of each other. They extend indefinitely, approaching but never touching their asymptotes. The asymptotes intersect at the center \((h, k)\), guiding the curvature of the hyperbola.

Completing the Square

Completing the square is a useful algebraic technique for transforming a quadratic expression into a perfect square trinomial. This method helps in finding the standard form of conic sections such as ellipses, parabolas, and hyperbolas. Here’s how it works:

• Given a quadratic expression, \(ax^2 + bx + c\), focus first on \(x\)-related terms.
• Rearrange and factor out the leading coefficient from \(x^2\) and \(x\) terms if necessary. For example, \(16(x^2 - 6x)\).
• Take the coefficient of \(x\), divide by two, and square it. For instance, divide \(-6\) by \(2\) to get \(-3\), square it to obtain \(9\).
• Add and subtract this square inside the expression, effectively maintaining the equality: \((x^2 - 6x + 9 - 9)\).
• Simplify to reveal a perfect square trinomial: \((x-3)^2\).

By completing the square in the original exercise, we reformat the problem into a recognizable form, allowing us to identify it as a hyperbola. This technique is invaluable for shifting from general quadratic forms to easily interpreted conic section equations.

Ellipse

An ellipse is another type of conic section, resembling an elongated circle. Its standard form equation is:

• \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]

Unlike the hyperbola, the ellipse has plus signs between the squared terms. Here,

• \((h, k)\) is the center of the ellipse,
• \(a\) and \(b\) represent the semi-major and semi-minor axes, respectively.

In general,

• If \(a > b\), the ellipse is stretched more along the \(x\)-axis.
• If \(b > a\), it is stretched along the \(y\)-axis.

Ellipses differ from hyperbolas as they are closed shapes and do not have asymptotes. They encapsulate a finite area, making them fitting models for planetary orbits and other similar natural phenomena.

Vertices and Foci

Vertices and foci are critical components when analyzing conic sections, specifically for ellipses and hyperbolas.

• **Vertices**: These are the points where the conic section intersects its major axis. In a hyperbola, they lie on the transverse axis—the line segment that passes through the center and vertices. The equation \((h \pm a, k)\) gives the vertices of a horizontally opening hyperbola and \((h, k \pm b)\) for a vertically opening one.
• **Foci**: These are specific points from which distances to any point on the hyperbola maintain a constant difference. For a hyperbola centered at \((h, k)\), the foci are located at \((h \pm c, k)\) where \(c\) is calculated using \(c^2 = a^2 + b^2\).

• In ellipses, the vertices are also determined by the distances along the major axis, which are \((h \pm a, k)\) for horizontal or \((h, k \pm b)\) for vertical axes.
• The foci of an ellipse lie on the major axis too, governed by \(c^2 = a^2 - b^2\).

Understanding these geometrical elements helps in sketching and interpreting the shapes, thus making conic sections more comprehensible.