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Chapter 7: Problem 47

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

Short Answer

Expert verified
The foci of the given ellipse are located at \((1+\sqrt{3}, -3)\) and \((1-\sqrt{3}, -3)\)

Step by step solution

01

Find the center of the ellipse

The center (h,k) of an ellipse equation is given by \((h,k)=(x,y)\) where \(x=h\) and \(y=k\). Substituting the values from the given equation, we get \(h=1\) and \(k=-3\). So, the center of the ellipse is (1, -3)
02

Find lengths of semi-major axis and semi-minor axis

The lengths of semi-major axis 'a' and semi-minor axis 'b' are equal to the square root of the denominators of \(x^2\) and \(y^2\) respectively. So here, the semi-major axis \(a=\sqrt{5}\) and the semi-minor axis \(b=\sqrt{2}\)
03

Apply the formula for the location of the foci

The foci are located at a distance of \(e=\sqrt{a^2-b^2}\) from the center along the major axis. Here, \(e=\sqrt{5-2}=\sqrt{3}\). Therefore, the foci are at \((h+e , k)\) and \((h-e , k)\) that is \((1+\sqrt{3},-3)\) and \((1-\sqrt{3},-3)\)
04

Draw the ellipse

Plot the center of the ellipse on the graph. Then, move a units along horizontal axes and b units along vertical axes to draw the ellipse. Plot the points corresponding to the Foci on the graph.

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Most popular questions from this chapter

Graph each ellipse and give the location of its foci. $$\frac{(x-4)^{2}}{9}+\frac{(y+2)^{2}}{25}=1$$

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

Use the center, vertices, and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. $$(y-2)^{2}-(x+3)^{2}=5$$

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

In Exercises \(1-18,\) graph each ellipse and locate the foci. $$\frac{x^{2}}{4}+\frac{y^{2}}{25}=1$$
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