Problem 21
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. $$ x^{2}-4 y^{2}-2 x+16 y=20 $$
Step-by-Step Solution
Verified Answer
The equation is a hyperbola with center (1, 2), vertices (1±√5, 2), and asymptotes y=2±(1/2)(x−1).
1Step 1: Group and Rearrange Terms
First, rearrange the equation by grouping the terms with \(x\) and \(y\) together. We have: \(x^2 - 2x - 4y^2 + 16y = 20\).
2Step 2: Complete the Square for X-Terms
For the \(x\)-terms \(x^2 - 2x\), complete the square. Take half of the coefficient of \(x\) (which is -2), square it to get 1, and add and subtract this inside the equation: \((x^2 - 2x + 1 - 1)\), resulting in: \((x - 1)^2 - 1\).
3Step 3: Complete the Square for Y-Terms
For the \(y\)-terms \(-4y^2 + 16y\), complete the square. Factor out \(-4\) to get \(-4(y^2 - 4y)\). Take half of the coefficient of \(y\) (after factoring, which is -4), square it to get 4, and add and subtract this inside the parentheses: \(-4((y - 2)^2 - 4)\), resulting in \(-4(y - 2)^2 + 16\).
4Step 4: Simplify and Set Equation
Substitute these back into the original equation: \((x - 1)^2 - 1 - 4(y - 2)^2 + 16 = 20\). Simplify to \((x - 1)^2 - 4(y - 2)^2 = 5\).
5Step 5: Identify the Conic Section
The equation \((x - 1)^2 - 4(y - 2)^2 = 5\) is of the form \((x - h)^2/a^2 - (y - k)^2/b^2 = 1\) which is a standard form of a hyperbola. Thus, it is a hyperbola.
6Step 6: Find Center and Characteristics
The center of the hyperbola is at \((h, k) = (1, 2)\). Since \(a^2 = 5\), \(b^2 = 5/4\). Calculate \(c = \sqrt{a^2 + b^2} = \sqrt{5 + 5/4} = \sqrt{25/4} = 5/2\). The foci are \((1 \pm \frac{5}{2}, 2)\). The vertices are \((1 \pm \sqrt{5}, 2)\).
7Step 7: Find Asymptotes
For a hyperbola centered at \((h, k)\) and vertices along the \(x\)-axis, the asymptotes are obtained by \(y = k \pm \frac{b}{a} (x-h)\). Here, \(b = \frac{\sqrt{5}}{2}, a = \sqrt{5}\), so the asymptotes are \(y = 2 \pm \frac{1}{2}(x-1)\).
8Step 8: Sketch the Graph
Graph the hyperbola by plotting the center at \((1, 2)\), the vertices and foci previously calculated, and draw the asymptotes to guide the shape of the hyperbola. The hyperbola will open left and right.
Key Concepts
HyperbolaCompleting the SquareGraphing ConicsAsymptotes
Hyperbola
A hyperbola is a fascinating type of conic section formed by the intersection of a double cone with a plane. Unlike ellipses or circles, hyperbolas have a distinctive shape that comprises two separate curves, called branches. These branches open in opposite directions and are mirror images of each other.
The equation for a hyperbola is usually presented in the standard form \((x - h)^2/a^2 - (y - k)^2/b^2 = 1\). This form indicates the location of the hyperbola's center at \((h, k)\), and reveals the orientation of its main axis. The key characteristics of a hyperbola include:
The equation for a hyperbola is usually presented in the standard form \((x - h)^2/a^2 - (y - k)^2/b^2 = 1\). This form indicates the location of the hyperbola's center at \((h, k)\), and reveals the orientation of its main axis. The key characteristics of a hyperbola include:
- Center: The midpoint \((h, k)\) between the focal points.
- Vertices: The nearest points to the center on each branch, lying along the axis.
- Foci: Located along the main axis, further away from the center than the vertices.
- Asymptotes: Lines that the hyperbola approaches as it extends further from the center.
Completing the Square
Completing the square is an essential algebraic technique used to rewrite quadratic expressions as perfect squares. This method is particularly helpful for transforming conic section equations into their standard forms.
To complete the square, especially when dealing with the equation like \(x^2 - 2x - 4y^2 + 16y = 20\), you follow these steps:
To complete the square, especially when dealing with the equation like \(x^2 - 2x - 4y^2 + 16y = 20\), you follow these steps:
- Reorganize the terms: Group terms that involve the same variable, like \(x\) or \(y\).
- Find the square completion: For each variable group, take half the linear coefficient, square it, then add and subtract it within the expression.
- Adjust and Simplify: Make adjustments by adding or subtracting constant terms to both sides of the equation to balance it.
Graphing Conics
Graphing conics like hyperbolas involves a series of steps to accurately depict their unique shapes. After completing the square and determining the conic's type and properties, these steps guide you:
- Identify the Center: From the standard form \((x - h)^2/a^2 - (y - k)^2/b^2 = 1\), locate the hyperbola's center \((h, k)\).
- Plot the Vertices and Foci: Calculate these key points using the derived values from the equation to establish the hyperbola's framework.
- Sketch the Asymptotes: The hyperbola's asymptotes serve as lines that guide the open branches of the curve.
- Draw the Hyperbola: Sketch each branch curving inward towards the center but never touching the asymptotes.
Asymptotes
Asymptotes are critical elements in graphing hyperbolas, characterizing the behavior of the curves as they extend toward infinity.
Asymptotes are not part of the hyperbola but rather serve as invisible boundaries that the hyperbola approaches without ever touching. For any hyperbola with an equation like \((x - h)^2/a^2 - (y - k)^2/b^2 = 1\), the asymptotes are derived based on:
Asymptotes are not part of the hyperbola but rather serve as invisible boundaries that the hyperbola approaches without ever touching. For any hyperbola with an equation like \((x - h)^2/a^2 - (y - k)^2/b^2 = 1\), the asymptotes are derived based on:
- Equation: The asymptotes for horizontally opening hyperbolas are expressed as \(y = k \pm \frac{b}{a}(x-h)\).
- Slope: The slope is \(\frac{b}{a}\) for the given hyperbola's orientation, determining how steeply the lines are drawn.
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