Chapter 11: Problem 32
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. $$x^{2}-y^{2}=10(x-y)+1$$
Short Answer
Step by step solution
Rearrange the Equation
Complete the Square for x-terms
Complete the Square for y-terms
Substitute back into Equation
Identify the Conic Section
Determine Center, Vertices, and Foci
Sketch the Hyperbola
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Completing the Square
Completing the square involves manipulating a quadratic expression so that it becomes a perfect square trinomial. For a term like \( x^2 - bx \), we add and subtract \( \left(\frac{b}{2}\right)^2 \) to create a perfect square. Here's how it works in practice:
- Identify the \'x\' and \'y\' groups separately in your equation.
- For each group, determine the value necessary to complete the square.
- Add and subtract this value within each group to maintain equation balance.
Hyperbola
Mathematically, the equation of a hyperbola in its standard form involves a subtraction between two squared terms, and it looks like this: \[(x - h)^2/a^2 - (y - k)^2/b^2 = 1\]. Here, \((h, k)\) indicates the center of the hyperbola, and \(a\) and \(b\) represent distances that define the hyperbola's shape.
- A hyperbola has two separate branches.
- The minus sign indicates that the foci lie along the transverse axis.
- The equation's right side is always \(1\) when the hyperbola crosses the y-axis.
Vertices of a Hyperbola
To find the vertices coordinates in an equation of the form \((x - h)^2/a^2 - (y - k)^2/b^2 = 1\), you simply move \(a\) units in either direction from the center \((h, k)\) along the transverse axis. For our equation:
- The center is at \((5, 5)\).
- Since \(a=1\), the vertices are located one unit left and right of the center.
- Thus, the vertices are at (6, 5) and (4, 5).
Foci of a Hyperbola
To find the foci of a hyperbola given by the equation \((x - h)^2/a^2 - (y - k)^2/b^2 = 1\), use the following steps:
- Calculate the distance \(c\) from the center to each focus using the formula \( c = \sqrt{a^2 + b^2} \).
- For our problem, since both \(a\) and \(b\) equal 1, \(c = \sqrt{1^2 + 1^2} = \sqrt{2}\).
- The foci are positioned \(c\) units away from the center along the transverse axis.
- Thus, the foci of the hyperbola are \((5 + \sqrt{2}, 5)\) and \((5 - \sqrt{2}, 5)\).
Asymptotes of a Hyperbola
For a hyperbola with its standard equation \((x - h)^2/a^2 - (y - k)^2/b^2 = 1\), the asymptotes are found using the equation:\( y - k = \pm \frac{b}{a} (x - h) \). This describes two intersecting lines that pass through the center of the hyperbola at \((h, k)\). For our hyperbola:
- The center is \( (5, 5) \).
- The slopes \( \pm \frac{1}{1} = \pm 1 \) indicate 45-degree angles.
- So, the asymptotes are \(y - 5 = \pm 1(x - 5)\), simplifying to \(y = x\) and \(y = -x + 10\).