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Chapter 11: Problem 49

Complete the square to determine whether the graph of the equation is an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an cllipse, 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. $$x^{2}-5 y^{2}-2 x+20 y=44$$

Short Answer

Expert verified
The equation represents a horizontal hyperbola with center (1, 2), vertices at (-4, 2) and (6, 2), foci at (1卤鈭�30, 2), and asymptotes are y鈭�2=卤(1/5)(x鈭�1).

Step by step solution

01

Rearrange the Equation

Start by rewriting the given equation: \[ x^2 - 5y^2 - 2x + 20y = 44. \]Separate the terms involving \(x\) and \(y\): \[ (x^2 - 2x) - 5(y^2 - 4y) = 44. \]
02

Complete the Square for x

Take the expression \(x^2 - 2x\) and complete the square:1. Identify the coefficient of \(x\), which is \(-2\).2. Divide it by 2 and square it: \((-2/2)^2 = 1\).3. Add and subtract 1: \[ x^2 - 2x + 1 - 1 = (x-1)^2 - 1. \]
03

Complete the Square for y

Take the expression \(y^2 - 4y\) and complete the square:1. Identify the coefficient of \(y\), which is \(-4\).2. Divide it by 2 and square it: \((-4/2)^2 = 4\).3. Add and subtract 4 inside the parentheses:\[ y^2 - 4y + 4 - 4 = (y-2)^2 - 4. \]Multiply by \(-5\) to maintain equation balance:\[ -5[(y-2)^2 - 4] = -5(y-2)^2 + 20. \]
04

Rewrite the Equation

Substitute the completed square forms from Steps 2 and 3 back into the equation:\[ (x-1)^2 - 1 - 5(y-2)^2 + 20 = 44. \]Simplify to obtain:\[ (x-1)^2 - 5(y-2)^2 = 25. \]
05

Identify the Conic Section

The equation \(\frac{(x-1)^2}{25} - \frac{(y-2)^2}{5} = 1\) represents a hyperbola.The standard form of a hyperbola is \((x - h)^2/a^2 - (y - k)^2/b^2 = 1\), indicating a horizontal hyperbola.
06

Find the Center, Foci, and Vertices

Identify the center, vertices, and foci:- **Center**: \((1, 2)\).- **Vertices**: Distance \( \pm a = \sqrt{25} = 5 \) from the center along the x-axis, yielding vertices at \((1 - 5, 2) = (-4, 2)\) and \((1 + 5, 2) = (6, 2)\).- **Foci**: Distance \( c = \sqrt{25 + 5} = \sqrt{30} \) from the center along the x-axis, giving foci at \( (1 - \sqrt{30}, 2) \) and \( (1 + \sqrt{30}, 2) \).
07

Find the Asymptotes

The slopes of the asymptotes for a horizontal hyperbola are \( \pm \frac{b}{a} = \pm \frac{ ext{1 (source from denominator of y part)}}{ ext{5}}\) giving:- Asymptotes: \( y - 2 = \pm \frac{1}{5}(x-1) \, .\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hyperbola
A hyperbola is a type of conic section that can be visualized as an open curve formed by intersecting a plane with a double cone. Unlike ellipses and circles, a hyperbola has two distinct branches that extend infinitely. These curves are defined by their unique algebraic equation format, which can be distinguished from that of ellipses and parabolas by the subtraction of squares.
  • Standard form: \((x - h)^2/a^2 - (y - k)^2/b^2 = 1\).
  • The variables \(a\) and \(b\) are constants that define the shape and orientation of the hyperbola.
  • Hyperbolas can be oriented horizontally or vertically, depending on which term is positive.
This exercise demonstrates how the method of completing the square helps identify a hyperbola by rearranging the given equation into the standard form.
Complete the Square
Completing the square is a mathematical technique used to simplify quadratic expressions, making it easier to analyze and solve associated equations. This process involves rearranging a quadratic expression to form a perfect square trinomial, which then can be rewritten in its factored form.
Follow these steps to complete the square:
  • Identify the coefficient of the linear term (the term without a square).
  • Divide this coefficient by 2 and square the result.
  • Add and subtract this squared value within the expression to maintain equality.
This technique allows you to transform equations into a standard form, simplifying the process of identifying conic sections like hyperbolas. In this exercise, completing the square for \(x\) and \(y\) enabled the identification of the hyperbola.
Center of a Hyperbola
The center of a hyperbola is a critical point that helps define the graph and is used as a reference for other properties like vertices and foci. It's the midpoint between the vertices and lies on the intersection of the hyperbola's conjugate axis and transverse axis.
  • In a standard hyperbola equation, the center is represented by \( (h, k) \).
  • The center is a starting point for measuring distances to the vertices and foci.
In this exercise, by completing the square, the equation was rearranged to identify the center at \( (1, 2) \). This serves as the origin for plotting other key elements of the hyperbola.
Vertices of a Hyperbola
Vertices are essential features of any hyperbola. They define the closest points of each branch to the hyperbola鈥檚 center and are instrumental in visualizing the curve's overall shape.
  • For a horizontal hyperbola, vertices lie on the transverse axis at \( (h \pm a, k) \).
  • The distance \(a\) is derived from the equation's denominator beneath the \(x\) term in the standard form.
In our example, the vertices were calculated to be \((-4, 2)\) and \( (6, 2) \), based on the center at \( (1, 2) \) with \(a = 5\).
Foci of a Hyperbola
Foci are integral to the definition of a hyperbola. These are fixed points, and the hyperbola itself is defined mathematically by the distances to these points. The difference between distances from any point on the hyperbola to each focus remains constant.
  • The formula for the distance from the center to each focus is \(c = \sqrt{a^2 + b^2}\).
  • Foci for a horizontal hyperbola are located at \( (h \pm c, k) \).
In this problem, the foci were calculated as \( (1 - \sqrt{30}, 2) \) and \( (1 + \sqrt{30}, 2) \), using the center at \( (1, 2) \). The distance \(c\) was derived from calculated values of \(a\) and \(b\).
Asymptotes of a Hyperbola
Asymptotes are diagonal lines that the branches of a hyperbola approach but never reach. They serve as a visual tool for sketching the hyperbola, indicating its direction and relative steepness.
  • For a horizontal hyperbola, asymptotes are given by the slopes \( \pm b/a \).
  • The equations for these lines are \(y - k = \pm (b/a)(x - h)\).
In the example provided, the asymptotes were defined with the equations \(y - 2 = \pm \frac{1}{5}(x - 1) \), guiding the drawing of the hyperbola based on given center, \(a\), and \(b\) values.

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Most popular questions from this chapter

Use a graphing device to graph the parabola. $$4 x+y^{2}=0$$

Finding the Equation of an Ellipse Find an equation for the ellipse that satisfies the given conditions. Foci: \((\pm 5,0),\) length of major axis: 12

Finding the Equation of an Ellipse Find an equation for the ellipse that satisfies the given conditions. Foci: \((0, \pm 2),\) length of minor axis: 6

Find the intersection points of the pair of ellipses. Sketch the graphs of each pair of equations on the same coordinate axes, and label the points of intersection. $$\left\\{\begin{aligned}25 x^{2}+144 y^{2} &=3600 \\\144 x^{2}+25 y^{2} &=3600\end{aligned}\right.$$

The ancillary circle of an ellipse is the circle with radius equal to half the length of the minor axis and center the same as the ellipse (see the figure). The ancillary circle is thus the largest circle that can fit within an ellipse. (a) Find an equation for the ancillary circle of the ellipse $$x^{2}+4 y^{2}=16$$ (b) For the ellipse and ancillary circle of part (a), show that if \((s, t)\) is a point on the ancillary circle, then \((2 s, t)\) is a point on the ellipse. (IMAGE CAN'T COPY)
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