Problem 25 - Algebra and Trigonometry

Question

\(23-34\) 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 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. $$ x^{2}-4 y^{2}-2 x+16 y=20 $$

Step-by-Step Solution

Answer

The equation represents a hyperbola with center (1, 2), vertices (1±√19, 2), foci (1±√(95/4), 2), and asymptotes at y = 1/2(x-1) + 2 and y = -1/2(x-1) + 2.

Step 1: Group and Rearrange Terms

First, collect the terms involving \(x\) and \(y\). So, rewrite the equation as: \[ x^2 - 2x - 4y^2 + 16y = 20 \] Group the \(x\)-terms and \(y\)-terms separately: \[ (x^2 - 2x) - (4y^2 - 16y) = 20 \]

Step 2: Complete the Square for x-terms

For the \(x\)-terms: Take \(x^2 - 2x\) and complete the square. 1. Take half of the coefficient of \(x\), which is \(-1\), and square it: \((-1)^2 = 1\). 2. Add and subtract inside the equation: \[ (x^2 - 2x + 1 - 1) - (4y^2 - 16y) = 20 \] 3. Factor the perfect square: \[ (x-1)^2 - (4y^2 - 16y) = 20 - 1 \] Simplify to: \[ (x-1)^2 - (4y^2 - 16y) = 19 \]

Step 3: Complete the Square for y-terms

For the \(y\)-terms: Take \(4y^2 - 16y\) and complete the square. 1. Factor out \(4\): \[ -4(y^2 - 4y) \]2. Take half of the coefficient of \(y\), which is \(-2\), and square it: \((-2)^2 = 4\).3. Add and subtract inside of \(y\)-part: \[ -4(y^2 - 4y + 4 - 4) \]4. Factor the perfect square and simplify further:\[ -4((y-2)^2 - 4) = -4(y-2)^2 + 16 \]Combine with earlier expression:\[ (x-1)^2 - 4(y-2)^2 = 19 \]

Step 4: Identify the Conic Type

The equation obtained is: \[ (x-1)^2 - 4(y-2)^2 = 19 \]This is the standard form of a hyperbola. We recognize it by the subtraction between terms and no single squared term indicating a parabola or an ellipse.

Step 5: Find Center, Foci, and Vertices of the Hyperbola

Center: \((h, k) = (1, 2)\).Vertices are given by solving: \[ x-1 = \pm \sqrt{19} \rightarrow (1 \pm \sqrt{19}, 2) \].Foci are given with formula \(c = \sqrt{a^2 + b^2}\) where \(a^2 = 19\) and \(b^2 = \frac{19}{4}\).Thus, \(c = \sqrt{19 + \frac{19}{4}} = \sqrt{\frac{95}{4}}\).Points for foci along the x-axis calculated as \((1 \pm \sqrt{\frac{95}{4}}, 2)\).

Step 6: Determine and Write Asymptotes

Asymptotes formula for horizontal hyperbola are: \[ y-k = \pm \frac{b}{a}(x-h) \] where \(a = \sqrt{19}\) and \(b = \frac{\sqrt{19}}{2}\).Calculating leads to: \[ y - 2 = \pm \frac{1}{2}(x - 1) \], thus equations of asymptotes are: \[ y = \frac{1}{2}(x - 1) + 2 \] and \[ y = -\frac{1}{2}(x - 1) + 2 \].

Key Concepts

Hyperbola

A hyperbola is one of the four types of conic sections, which also includes circles, ellipses, and parabolas. Conic sections are shapes that can be obtained by slicing a cone at different angles. Hyperbolas have a distinctive open shape; they consist of two disconnected curves called branches. The general equation for a hyperbola looks like this: \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \] or \[ \frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} = 1 \] Notice the subtraction between the terms; that's a key identifier of a hyperbola. In this exercise, after completing the square, the equation \[ (x-1)^2 - 4(y-2)^2 = 19 \] was obtained, identifying it as a hyperbola. The subtraction between \((x-1)^2\) and \(-4(y-2)^2\) indicates the hyperbola's structure, separating it from parabolas or ellipses.

Completing the Square

Completing the square is an essential algebraic technique used to simplify quadratic equations and recognize their forms more easily. In the context of conic sections, this method helps transform quadratic equations into a recognizable standard form, such as the one for hyperbolas. To complete the square:

• Rearrange the quadratic terms together.
• Take half of the linear term's coefficient of a squared variable, square it, and add it to both sides of the equation.
• Factor the newly formed perfect square trinomial.

In this exercise, the steps started with the equation \(x^2 - 2x - 4y^2 + 16y = 20\). For \(x\)-terms, \((x-1)^2\) was formed, and for \(y\)-terms, \(-4(y-2)^2\) was factored. Completing the square for both variables allowed the equation to be rewritten in a form where you can easily identify the hyperbola.

Center of Hyperbola

The center of a hyperbola is a point that lies directly in the middle of its two branches. This point acts as a reference for locating other key features like the vertices and foci. For hyperbolas, the center point is noted in the equation \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \] as \((h, k)\). Here, it was determined after completing the square that the center is at (1, 2). Finding the center is simple once the equation is in standard form; it reveals itself as the fixed coordinates shifting the graph's origin. In this context, no computations are needed beyond recognizing \((h, k)\) in the rewritten equation.

Equations of Asymptotes

Asymptotes are imaginary lines that the branches of a hyperbola approach but never touch. They guide the hyperbola's curvature and provide insight into its orientation, whether horizontally or vertically aligned. The formula for determining asymptotes of a hyperbola can be expressed as: \[ y-k = \pm \frac{b}{a}(x-h) \] For this exercise, after simplifying and computing from \[ \frac{a}{b} \] where \(a = \sqrt{19}\) and \(b = \frac{\sqrt{19}}{2}\), the equations of the asymptotes were determined: \[ y = \frac{1}{2}(x - 1) + 2 \] and \[ y = -\frac{1}{2}(x - 1) + 2 \]. These equations tell you how sharply the hyperbola's branches spread away from the center, giving the graph direction and orientation.