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Chapter 13: Problem 60

For the following equations of hyperbolas, complete the square, if necessary, and write in standard form. Find the center, the vertices, and the asymptotes. Then graph the hyperbola. \(\frac{(x-2)^{2}}{9}-\frac{(y-1)^{2}}{4}=1\)

Short Answer

Expert verified
Center: \((2, 1)\), Vertices: \((-1, 1)\) and \((5, 1)\), Asymptotes: \(y = 1 \pm \frac{2}{3}(x - 2)\).

Step by step solution

01

Identify the Form

Recognize that the given equation is already in its standard form for a hyperbola: \[\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1\]
02

Determine the Center

Compare the given equation \( \frac{(x-2)^{2}}{9}-\frac{(y-1)^{2}}{4}=1 \) with the standard form. The center \((h, k)\) can be identified directly from the equation as \((2, 1)\).
03

Find the Vertices

For the horizontal hyperbola, the vertices are a distance \(a\) from the center along the x-axis. Here, \(a^{2} = 9\) thus \(a = 3\). Vertices: \((2-3, 1)\) and \((2+3, 1)\) which simplifies to \((-1, 1)\) and \((5, 1)\).
04

Find the Asymptotes

The asymptotes for a hyperbola in this form are given by the equations \(y = k \pm \frac{b}{a}(x-h)\). Here, \(b^{2} = 4\) so \(b = 2\). Therefore, the asymptotes are: \[y = 1 \pm \frac{2}{3}(x-2)\]
05

Graph the Hyperbola

Plot the center at \((2, 1)\). Plot the vertices at \((-1, 1)\) and \((5, 1)\). Use the equations of the asymptotes to draw the asymptotes lines: \(y = 1 + \frac{2}{3}(x-2)\) and \(y = 1 - \frac{2}{3}(x-2)\). Finally, draw the hyperbola opening left and right passing through the vertices.

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Key Concepts

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

standard form of hyperbola
To solve hyperbola equations, it's crucial to recognize the standard form. A standard equation for a hyperbola opens horizontally: \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \] and one for a hyperbola that opens vertically is: \[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \]. Here:
  • h and k represent the coordinates of the center of the hyperbola.
  • a is the distance from the center to the vertices.
  • b helps find the equations of the asymptotes and the shape of the hyperbola.
Starting with the correct form simplifies the problem immensely.
center of hyperbola
The center of a hyperbola is straightforward to identify once the equation is in standard form. For \[ \frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1 \], the center is at \((h,k)\).Given the equation in the problem, \[ \frac{(x-2)^{2}}{9}-\frac{(y-1)^{2}}{4}=1 \], we directly see that h=2 and k=1, hence the center is \((2,1)\). This center is pivotal for graphing and understanding the hyperbola鈥檚 layout.
vertices of hyperbola
Vertices are the points where the hyperbola intersects its principal axis. To find them:
  • Identify a, where a is derived from a^2 in the standard form equation. For our equation, a^2 = 9, so a = 3.
  • The vertices are a units from the center along the x-axis if the hyperbola opens horizontally, or along the y-axis if it opens vertically. Since \[ \frac{(x-2)^{2}}{9}-\frac{(y-1)^{2}}{4}=1 \] opens horizontally, we find the vertices by adding and subtracting 3 from the x-coordinate of the center: (2-3, 1) and (2+3, 1), which are (-1,1) and (5,1).
Vertices are crucial as they mark the ends of the hyperbola鈥檚 transverse axis.
asymptotes of hyperbola
Asymptotes are the lines that the hyperbola approaches but never touches. They define the shape of the hyperbola. For the horizontal hyperbola in the standard form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), the asymptotes are given by \( y = k \pm \frac{b}{a}(x-h) \).
  • Identify b, which comes from b^2 in the equation. For our hyperbola, b^2=4, so b=2.
  • Substitute h, k, a, and b into the asymptote equations: \( y = 1 \pm \frac{2}{3}(x-2) \).
These lines help guide the shape of your hyperbola on the graph.
graphing hyperbolas
When graphing hyperbolas, combine all the elements. Follow these steps:
  • Plot the center of the hyperbola at (2,1).
  • Mark the vertices at (-1,1) and (5,1).
  • Draw the asymptotes using \( y = 1 + \frac{2}{3}(x-2) \) and \( y = 1 - \frac{2}{3}(x-2) \).
  • Sketch the hyperbola by drawing two curves passing through the vertices and approaching the asymptotes, opening left and right.
With all these elements, you create an accurate graph of the hyperbola. Practice these steps to become proficient in quickly graphing hyperbolas.

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Most popular questions from this chapter

Match the equation with the center or vertex of its graph, listed in the column on the right. a) Vertex: \((-2,5)\) b) Vertex: \((5,-2)\) c) Vertex: \((2,-5)\) d) Vertex: \((-5,2)\) e) Center: \((-2,5)\) f) Center: \((2,-5)\) g) Center: \((5,-2)\) h) Center: \((-5,2)\) $$(x+5)^{2}+(y-2)^{2}=9$$

Match the equation with the center or vertex of its graph, listed in the column on the right. a) Vertex: \((-2,5)\) b) Vertex: \((5,-2)\) c) Vertex: \((2,-5)\) d) Vertex: \((-5,2)\) e) Center: \((-2,5)\) f) Center: \((2,-5)\) g) Center: \((5,-2)\) h) Center: \((-5,2)\) $$x=(y-2)^{2}-5$$

Fireside Castings finds that the total revenue \(R\) from the sale of \(x\) units of railing is given by $$ R=100 x+x^{2} $$ Fireside also finds that the total cost \(C\) of producing \(x\) units of the same product is given by $$ C=80 x+1500 $$ A break-even point is a value of \(x\) for which total revenue is the same as total cost; that is, \(R=C\) How many units must be sold in order to break even?

Tile Design. The Old World tile company wants to make a new rectangular tile that has a perimeter of 6 in. and a diagonal of length \(\sqrt{5}\) in. What should the dimensions of the tile be? PICTURE NOT AVAILABLE

Let \((-c, 0)\) and \((c, 0)\) be the foci of an ellipse. Any point \(P(x, y)\) is on the ellipse if the sum of the distances from the foci to \(P\) is some constant. Use \(2 a\) to represent this constant. A) Show that an equation for the ellipse is given by \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}-c^{2}}=1\) B) Substitute \(b^{2}\) for \(a^{2}-c^{2}\) to get standard form.
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