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Chapter 7: Problem 39

Use the center, vertices, and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. $$(x-3)^{2}-4(y+3)^{2}=4$$

Short Answer

Expert verified
The center of the hyperbola is at (3, -3). The vertices are at (1, -3) and (5, -3). The foci are at (3 - 2鈭�2, -3) and (3 + 2鈭�2, -3). Equations of the asymptotes are \(y=x-6\) and \(y=-x\).

Step by step solution

01

Identify the Center

The center of the hyperbola is given by the values (h, k) where h is the value opposite the x term and k is the value opposite the y term. In the given equation, the center is at (3, -3).
02

Identify a and b

a is the coefficient of x squared and b is the coefficient of y squared. In this case, 'a' is square root of 4 (the number accompanying the x squared term) that equals 2. And 'b' is square root of 4 (the number accompanying the y squared term) that equals 2.
03

Identify the Vertices

Vertices are \((h\pm a, k)\) for horizontal hyperbola. Here, this equates to \((3\pm2 , -3)\), so the vertices are at (1, -3) and (5, -3).
04

Determine the Foci

Foci are \((h\pm c, k)\), where \(c=\sqrt{a^2+b^2}\). Compute this for c to get \(c=\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}\). Substituting these values gives the foci - (3 - 2鈭�2, -3) and (3 + 2鈭�2, -3).
05

Find the equations of the asymptotes

The asymptotes of a hyperbola are given by \(y = k \pm ((b/a)(x-h))\). Substituting the values, we get: \(y=-3\pm ((2/2)*(x-3)) =-3\pm(x-3)\). Thus, the equations of the asymptotes are \(y=x-6\) and \(y=-x\).

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Most popular questions from this chapter

In Exercises 43-50, convert each equation to standard form by completing the square on \(x\) and \(y .\) Then graph the hyperbola. Locate the foci and find the equations of the asymptotes. $$x^{2}-y^{2}-2 x-4 y-4=0$$

Convert each equation to standard form by completing the square on \(x\) and \(y .\) Then graph the hyperbola. Locate the foci and find the equations of the asymptotes. $$9 y^{2}-4 x^{2}-18 y+24 x-63=0$$

Convert each equation to standard form by completing the square on \(x\) and \(y .\) Then graph the hyperbola. Locate the foci and find the equations of the asymptotes. $$4 x^{2}-25 y^{2}-32 x+164=0$$

In Exercises \(1-18,\) graph each ellipse and locate the foci. $$\frac{x^{2}}{49}+\frac{y^{2}}{36}=1$$

Use the center, vertices, and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. $$(x+3)^{2}-9(y-4)^{2}=9$$
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