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Chapter 8: Problem 40

For the following exercises, sketch a graph of the hyperbola, labeling vertices and foci. $$ -x^{2}+8 x+4 y^{2}-40 y+88=0 $$

Short Answer

Expert verified
Vertices: (2, 5) and (6, 5). Foci: (4卤鈭�5, 5).

Step by step solution

01

Rewrite the Equation

First, let's rewrite the given equation by grouping the terms related to \(x\) and \(y\). The given equation is:\[-x^2 + 8x + 4y^2 - 40y + 88 = 0\]We'll rearrange it as:\[-(x^2 - 8x) + 4(y^2 - 10y) = -88\].
02

Complete the Square

Next, we complete the square for both the \(x\) and \(y\) terms. For \(-x^2 + 8x\): Complete the square:\[-(x^2 - 8x + 16 - 16) = -(x^2 - 8x + 16) + 16 = -(x-4)^2 + 16\].For \(4y^2 - 40y\):Factor out the 4:\[4(y^2 - 10y + 25 - 25) = 4((y-5)^2 - 25) = 4(y-5)^2 - 100\].
03

Redefine the Equation

Now substitute the completed square forms back into the equation:\[-(x-4)^2 + 16 + 4(y-5)^2 - 100 = -88\].Simplify the constant terms:\[-(x-4)^2 + 4(y-5)^2 - 84 = -88\]\[-(x-4)^2 + 4(y-5)^2 = -4\].
04

Divide to Simplify

Divide through by -4 to express the equation in the form of a hyperbola:\[\frac{(x-4)^2}{4} - (y-5)^2 = 1\].This can be rewritten as:\[\frac{(x-4)^2}{4} - \frac{(y-5)^2}{1} = 1\].
05

Identify the Hyperbola Features

From the equation \[\frac{(x-4)^2}{4} - \frac{(y-5)^2}{1} = 1\], we can identify the hyperbola's features:- Center: \((h, k) = (4, 5)\).- Vertices: Since \(a^2 = 4\), we have \(a = 2\), so vertices are at \((4卤2, 5)\) or \((2, 5)\) and \((6, 5)\).- Foci: Calculate using \(c = \sqrt{a^2 + b^2} = \sqrt{4 + 1} = \sqrt{5}\), so foci are \((4卤\sqrt{5}, 5)\).
06

Sketch the Hyperbola

With the center \((4, 5)\), vertices \((2, 5)\) and \((6, 5)\), and foci \((4卤\sqrt{5}, 5)\), sketch the horizontal hyperbola on a graph.- Draw the transverse axis through the vertices.- Mark the vertices and foci on the graph.- Sketch the asymptotes, which depend on slopes derived from \(b/a = 1/2\), passing through the center.- Draw the hyperbola approaching these asymptotes.

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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 is a technique used to transform a quadratic expression into a perfect square trinomial. This helps in rewriting a hyperbola equation so it's easier to understand the structure and characteristics of the hyperbola.
To complete the square, you need to follow these steps for each of the variable terms:
  • Identify the coefficient of the linear term.
  • Divide this coefficient by 2.
  • Square the result.
  • Add and subtract this square inside the equation to maintain the equality.
For instance, in the term \(x^2 - 8x\), note the following:
  • Linear coefficient is \(8\).
  • Half of it is \(4\), and its square is \(16\).
  • Rewrite as \(x^2 - 8x + 16 - 16\).
  • This simplifies to \((x-4)^2 - 16\).
This process changes the form of the equation, making it easier to identify features like the center and vertices of a hyperbola.
Vertices of Hyperbola
Vertices of a hyperbola are key points that help define its shape, located on the major axis of the hyperbola. Given a hyperbola in the standard form, vertices are determined by the values of \(a\) (the distance from the center to a vertex on the transverse axis).
  • For a horizontal hyperbola \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), vertices are \((h\pm a, k)\).
  • For the given equation, \(a^2 = 4\), so \(a = 2\).
  • With the center at \((4,5)\), vertices are \((4\pm2, 5) = (2, 5)\) and \((6, 5)\).
Vertices define the width of the hyperbola and indicate where it will stretch along the main direction. Identifying these points helps in sketching the hyperbola accurately.
Foci of Hyperbola
The foci (plural of focus) are two points inside each branch of a hyperbola. They are crucial for defining the nature of hyperbolas, influencing their width and openness. The distance from the center to each focus is \(c\), calculated using the relationship \(c = \sqrt{a^2 + b^2}\).
  • For \(a^2 = 4\) and \(b^2 = 1\), \(c = \sqrt{4 + 1} = \sqrt{5}\).
  • For the equation \({\frac{(x-4)^2}{4} - \frac{(y-5)^2}{1} = 1}\), foci reside on the transverse axis.
  • Thus, the foci are at \( (h\pm\sqrt{5}, k)\), or \((4\pm\sqrt{5}, 5)\).
Plotting the foci allows for setting up the asymptotes and understanding how the hyperbola approaches infinity.
Center of Hyperbola
The center of a hyperbola is a critical point from which the hyperbola is symmetrically stretched. It is akin to the midpoint between its vertices and foci.
  • The general form of a hyperbola's equation reveals the center \((h, k)\) from the quadratic components' completed square format.
  • In this hyperbola, transformed to \({\frac{(x-4)^2}{4} - \frac{(y-5)^2}{1} = 1}\), the center is clearly \((4,5)\).
Understanding the center's location is essential as it affects the positioning of all main features of the hyperbola. From here, you can accurately sketch the hyperbola, its asymptotes, and notable lines.

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Most popular questions from this chapter

For the following exercises, graph the given conic section. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse, label the vertices and foci. If it is a hyperbola, label the vertices and foci. $$ r=\frac{2}{3+3 \sin \theta} $$

For the following exercises, assume an object enters our solar system and we want to graph its path on a coordinate system with the sun at the origin and the \(x\) -axis as the axis of symmetry for the object's path. Give the equation of the flight path of each object using the given information. The object enters along a path approximated by the line \(y=3 x-9\) and passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line \(y=-3 x+9\).

For the following exercises, identify the conic with a focus at the origin, and then give the directrix and eccentricity. $$ r=\frac{3}{10+10 \cos \theta} $$

Graph the given conic section. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse, label the vertices and foci. If it is a hyperbola, label the vertices and foci. $$ r(6-4 \cos \theta)=5 $$

For the following exercises, identify the conic with a focus at the origin, and then give the directrix and eccentricity. $$ r(1-\cos \theta)=3 $$
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