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Chapter 7: Problem 28

Find the standard form of the equation of each ellipse satisfying the given conditions. $$\text { Foci: }(0,-3),(0,3) ; \text { vertices: }(0,-4),(0,4)$$

Short Answer

Expert verified
The standard form of the equation of the ellipse is \( \frac{x^2}{7} + \frac{y^2}{16} = 1 \)

Step by step solution

01

Determine values of a, b, and c

For the given ellipse, the distance from the center to a vertex (a) is 4 and the distance from the center to the foci (c) is 3. Hence, \(a=4\) and \(c=3\). The distance \(b\) can be calculated from the relationship between a, b, and c in an ellipse, which is \(a^2 = b^2+c^2\). Substituting a=4 and c=3 into this equation gives \(b^2 = a^2 - c^2 = 4^2 - 3^2 = 7\).
02

Write down the standard form of the ellipse equation

The standard form of an ellipse centered at the origin with vertical major axis is \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \). Substituting \(a=4\) and \(b = \sqrt{7}\) gives the equation of the ellipse as \( \frac{x^2}{7} + \frac{y^2}{16} = 1 \)

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Most popular questions from this chapter

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 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. $$x^{2}-y^{2}-2 x-4 y-4=0$$

Use a graphing utility to graph \(\frac{x^{2}}{4}-\frac{y^{2}}{9}=0 .\) Is the graph a hyperbola? In general, what is the graph of \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=0 ?\)

Graph each ellipse and give the location of its foci. $$\frac{(x-1)^{2}}{2}+\frac{(y+3)^{2}}{5}=1$$

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. $$4 x^{2}-9 y^{2}-16 x+54 y-101=0$$
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