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Chapter 7: Problem 50

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

Short Answer

Expert verified
The ellipse has its center at (-4,-3), lengths of the semi-major and semi-minor axes are 1 and 6 respectively, and the foci are located at (-4, -3+sqrt(35)) and (-4, -3-sqrt(35)).

Step by step solution

01

Determine the center of the ellipse

The center of the ellipse is given by the point (h,k). In the equation, h is subtracted from x and k is subtracted from y. However our equation, \(36(x+4)^{2} + (y+3)^{2} = 36\), has x and y added with numbers. So, the center will be the point at which these are 'negatives'. Hence, the center of the ellipse is at (-4,-3).
02

Determine the lengths of the semi-major and semi-minor axes

The general form of the equation of an ellipse is \(A(x-h)^{2} + B(y-k)^{2} = r^{2}\), where \(r_{x}=\sqrt{r^{2}/A}\) is the length of the semi-major axis and \(r_{y}=\sqrt{r^{2}/B}\) is the length of the semi-minor axis. Here, A=36, B=1 and r^2=36. So, \(r_{x}=\sqrt{36/36}\) gives 1 and \(r_{y}=\sqrt{36/1}\) gives 6. Therefore, the lengths of the semi-major and semi-minor axes are 1 and 6 respectively.
03

Determine the foci

The foci are located at a distance of \(c=\sqrt{r_{y}^{2}-r_{x}^{2}}\) from the center along the y-axis (since \(r_{y}>r_{x}\)). Substituting \(r_{y}=6\) and \(r_{x}=1\), we get \(c=\sqrt{6^{2}-1^{2}}=\sqrt{35}\). Therefore, the foci of the ellipse are at points (-4, -3+sqrt(35)) and (-4, -3-sqrt(35)).

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