Problem 19 - Precalculus: Mathematical 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. $$9 x^{2}-36 x+4 y^{2}=0$$

Step-by-Step Solution

Answer

The equation represents a hyperbola centered at (2, 0) with vertices at (0, 0) and (4, 0).

Step 1: Rearrange Terms

First, rearrange the terms in the given equation, grouping the x-terms and y-terms. The equation is \(9x^2 - 36x + 4y^2 = 0\). Rearrange it to separate the x and y components: \(9x^2 - 36x = -4y^2\).

Step 2: Complete the Square for X-Terms

Factor out the coefficient of \(x^2\) from the x-terms: \(9(x^2 - 4x) = -4y^2\). Now, complete the square inside the parentheses. The given expression is \(x^2 - 4x\). Take half of the coefficient of \(x\), square it: \((-4/2)^2 = 4\). Add and subtract this square inside the parentheses: \(x^2 - 4x + 4 - 4\). This becomes \((x - 2)^2 - 4\). Therefore, \(9((x - 2)^2 - 4) = -4y^2\).

Step 3: Simplify the Equation

Distribute the 9 across the completed square term: \(9(x - 2)^2 - 36 = -4y^2\). Rearrange the terms to get all terms on one side: \(9(x - 2)^2 - 4y^2 = 36\).

Step 4: Identify the Conic Section

Divide all terms by 36 to rewrite the equation: \(\frac{9(x-2)^2}{36} - \frac{4y^2}{36} = 1\), simplifying this gives \(\frac{(x-2)^2}{4} - \frac{y^2}{9} = 1\). This is the standard form of a hyperbola centered at \((h, k)\).

Step 5: Determine the Hyperbola Parameters

The hyperbola is centered at \((h, k) = (2, 0)\). Compare \(a^2\) and \(b^2\) terms: \(a^2 = 4\) and \(b^2 = 9\). Hence \(a = 2\) and \(b = 3\). The vertices are at \((h\pm a, k) = (2\pm2, 0) = (0, 0)\) and \((4, 0)\). For the foci, calculate \(c\) where \(c^2 = a^2 + b^2 = 13\), thus \(c = \sqrt{13}\). The foci are at \((h\pm c, k) = (2\pm\sqrt{13}, 0)\). Asymptotes are given by \(y = \pm\frac{3}{2}(x-2)\).

Step 6: Sketch the Hyperbola

To sketch, draw the rectangle formed by \(2a\) and \(2b\) (width of 4 and height of 6) centered at \((2,0)\). The hyperbola's branches open horizontally. Additionally, use the slopes from the asymptotes to guide the sketch.

Key Concepts

Completing the Square

Completing the square is a technique used to convert quadratic expressions into a perfect square trinomial. This method simplifies the analysis of conic sections, particularly when determining whether an equation represents an ellipse, parabola, or hyperbola. The process involves three simple steps:

• First, factor out the coefficient of the squared term, if necessary. For instance, given the expression involving \(x\), start by factoring out from \(9x^2 - 36x\).
• Next, take half the coefficient of the linear term, square it, and adjust the equation by adding and subtracting this value. For \(-4x\), half is \(-2\), and squaring yields \(4\). This adjustment transforms the expression to \((x - 2)^2 - 4\).
• Finally, distribute any factored coefficients back to simplify the expression while maintaining equality across the equation.

By completing the square, one can easily rewrite quadratic terms, which is particularly useful in identifying and graphing conic sections accurately.

Hyperbola

A hyperbola is one of the conic sections, defined as the locus of points where the difference of distances to two fixed points, called foci, is constant. Hyperbolas have distinctive features such as centers, axes, vertices, foci, and asymptotes.

The standard form of a hyperbola with a horizontal transverse axis is \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\]Here,

• \((h, k)\) is the center of the hyperbola.
• \(a\) determines the distance from the center to the vertices.
• \(b\) helps calculate the slope of the asymptotes and defines the width of the rectangle associated with the hyperbola.
• \(c = \sqrt{a^2 + b^2}\) gives the distance to the foci.

The vertices are found at \((h \pm a, k)\), providing the main axis of symmetry. Asymptotes have equations given by \(y = \pm\frac{b}{a}(x-h) + k\), guiding the opening of the hyperbola's branches in sketches.

Conic Section Parameters

Conic section parameters define the unique characteristics and equations for ellipses, parabolas, and hyperbolas. Understanding these parameters is crucial when analyzing and graphing these curves.

• Center: The fixed midpoint from which measurements are taken, labeled \((h, k)\) in standard equations.
• Vertices: Points where the conic intersects its principal axes. For hyperbolas and ellipses, they lay along the transverse axis.
• Foci: Fixed points used to define the curve. In parabolas, this is a single point, whereas hyperbolas and ellipses have two.
• Axes: For ellipses, these are the major and minor axes defining the direction of elongation. Hyperbolas have transverse and conjugate axes.

With the equation \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\], obtaining these parameters allows us to delineate specific characteristics such as the hyperbola's orientation and angle of approach through its asymptotes. Correctly determining these values aids in the accurate sketching and interpretation of conic sections.