Problem 25 - Algebra and Trigonometry

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) and vertices at (0,0) and (6,0). Foci are at (8,0) and (-2,0), with asymptotes \(y=\pm\frac{4}{3}(x-3)\).

Step 1: Reorganize Terms

Start by moving all the constant terms in the equation to the other side to prepare for completing the square.Equation: \(16x^2 - 96x - 9y^2 = -288\)

Step 2: Complete the Square for x-Terms

For the \(x\)-terms, factor out the common factor of 16 and complete the square:\[16(x^2 - 6x) = 16((x - 3)^2 - 9)\]So it becomes:\[16((x - 3)^2 - 9) = -288\]

Step 3: Simplify Completed Square for x-Terms

Distribute the 16:\[16(x-3)^2 - 144 = -288\]

Step 4: Address y-Terms

Since there's no need to complete the square for the \(y\) terms as they are ready to factor back:\[-9(y^2) = -9y^2\]

Step 5: Combine Completed Squares

Include the rearranged terms to set the completed square form:\[16(x-3)^2 - 9y^2 = -144\]

Step 6: Look for Simplification Opportunities

Divide the entire equation by \(-144\) to simplify and identify the conic form:\[\frac{(x-3)^2}{9} - \frac{y^2}{16} = 1\]

Step 7: Identify the Conic

Recognizing the equation \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) shows this is a hyperbola.Center = (3, 0), \(a^2 = 9\) and \(b^2 = 16\)

Step 8: Find Vertices and Foci

Calculate the points:Vertices at \((3 \pm 3, 0)\) which are \((0, 0)\) and \((6, 0)\).Calculate the foci using \(c = \sqrt{a^2 + b^2} = \sqrt{25} = 5\).Foci are at \((3 \pm 5, 0)\), giving \((8, 0)\) and \((-2, 0)\).

Step 9: Find Asymptotes

The slopes for the asymptotes are \(\pm \frac{b}{a} = \pm \frac{4}{3}\), giving equations:\(y = \pm \frac{4}{3}(x - 3)\)

Key Concepts

Completing the Square

Completing the square is a technique used to transform a quadratic equation into a perfect square trinomial. This method is essential when working with conic sections, as it helps to rewrite equations in their standardized forms. For the equation provided, we start by reorganizing and isolating the terms involving specific variables. In our example, the expression involves both \(x\)-terms and \(y\)-terms.

• We first focus on the \(x\)-terms: \(16x^2 - 96x\).
• Factor out the common factor (16 in this case).
• Reorganize these terms into a perfect square: \((x - 3)^2\).

Thus, transforming \(16(x^2 - 6x)\) into \(16((x-3)^2 - 9)\). Whence, distributing gives \(16(x-3)^2 - 144\). It is important to note that if \(y\)-terms also contain quadratic expressions not in the form of a perfect square, they should be completed utilizing a similar approach. Completing the square helps in identifying the nature of the conic by simplifying the equation into its standard form.

Hyperbola

A hyperbola is a type of conic section that occurs when the plane cuts through both nappes of the cone. The equation of a hyperbola takes the form: \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \] Here's what each symbol represents:

• \((h, k)\) is the center of the hyperbola.
• \(a^2\) and \(b^2\) denote the squared distances that determine the shape and orientation of the hyperbola.

For our example, after completing the square, we establish the equation in its hyperbola form:\[ \frac{(x-3)^2}{9} - \frac{y^2}{16} = 1 \] This indicates that the hyperbola is centered at the point \((3, 0)\). The terms related to \(x\) and \(y\) reveal the stretch and width of the hyperbola along both axes.

Vertices and Foci

Vertices and foci are critical to understanding the structure and characteristics of a hyperbola. The vertices of a hyperbola lie on the transverse axis, which is the line that passes through the center and intersects the two branches of the hyperbola.

• For the vertices, we calculate using \( a \) (where \( a = 3 \) in our example). The vertices are located at points \((h \, \pm \, a , k)\), resulting in \( (0, 0) \) and \( (6, 0) \).
• The foci are found using \( c \), where \( c = \sqrt{a^2 + b^2} = \sqrt{9 + 16} = 5 \).
• The foci points are \((h \, \pm \, c , k)\), resulting in \( (-2, 0) \) and \((8, 0) \).

The vertices indicate the points closest to the center that the hyperbola curves around, while the foci help in defining the hyperbola itself. The foci are typically further out along the axis of symmetry than the vertices.

Asymptotes

Hyperbolas are characterized by their asymptotes, which are straight lines that the curve approaches but never touches. Asymptotes provide a guide to how wide or narrow the hyperbola opens.For a hyperbola of the form \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \] The slopes of the asymptotes are given by \( \pm \frac{b}{a} \). In our equation, \( a^2 = 9 \) and \( b^2 = 16 \), hence the slopes are \( \pm \frac{4}{3} \).

• The equations of the asymptotes then become \( y = \frac{4}{3}(x-3) \) and \( y = -\frac{4}{3}(x-3) \).
• These lines guide the shape and direction of the hyperbola, crossing at the center, \((3,0)\).

Understanding asymptotes helps visualize and sketch hyperbolas accurately since these lines dictate the angle at which the hyperbola's arms extend to infinity.