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Chapter 7: Problem 35

Use the center, vertices, and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. $$\frac{(x+3)^{2}}{25}-\frac{y^{2}}{16}=1$$

Short Answer

Expert verified
The center of the hyperbola is (-3,0), the vertices are at (-8,0) and (2,0), and the foci are at (-3 卤 鈭�41,0). The equations of the asymptotes are \( y = \pm \frac{4}{5}(x+3) \).

Step by step solution

01

Identify the center of the hyperbola

The center of the hyperbola is given by the values inside the parentheses in the equation and with opposite signs. So, the center is at (-3,0).
02

Identify the vertices of the hyperbola

The vertices are located at a distance of \( \sqrt{25} = 5 \) units to the left and right of the center along the x-axis. So, the vertices are at (-8,0) and (2,0).
03

Identify and sketch the asymptotes

The asymptotes of the hyperbola are given by the lines passing through the center with slopes of \( \pm \sqrt{\frac{16}{25}} = \pm \frac{4}{5} \). The equations of the lines are \( y = \pm \frac{4}{5}(x+3) \). Plot these lines on the graph.
04

Sketch the hyperbola

Draw the hyperbola, making sure it approaches the asymptotes as \( x \) goes to \( \pm \infty \). The hyperbola passes through the vertices and opens to the left and right since the x term is positive in the equation.
05

Find the foci of the hyperbola

The distance from the center to each focus is given by \( \sqrt{25+16} = \sqrt{41} \) units. So the foci are at \( (-3 \pm \sqrt{41},0) \).

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Most popular questions from this chapter

Graph each ellipse and give the location of its foci. $$(x+3)^{2}+4(y-2)^{2}=16$$

Graph \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) and \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=-1\) in the same viewing rectangle for values of \(a^{2}\) and \(b^{2}\) of your choice. Describe the relationship between the two graphs.

Graph each ellipse and give the location of its foci. $$36(x+4)^{2}+(y+3)^{2}=36$$

Graph each ellipse and give the location of its foci. $$9(x-1)^{2}+4(y+3)^{2}=36$$

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. $$4 x^{2}-9 y^{2}+8 x-18 y-6=0$$
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