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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 center of the ellipse is at (-4, -3), with a semi-major axis of length 6 and a semi-minor axis of length 1. As such, the foci are found at (-4, -3+sqrt(35)) and (-4, -3-sqrt(35)).

Step by step solution

01

Identify the Center of the Ellipse

From the equation we can determine the center of the ellipse by looking at the values of h and k, which are the coefficients of x and y respectively. However, note that the standard equation considers positive (h,k), so if our equation is \((x+4)^{2}/1+(y+3)^{2}/36=1\), the center of the ellipse is at (-4,-3).
02

Determine the Length of the Semi-Major and Semi-Minor Axes

The a value is found under y in the equation, which is 6 (the square root of 36). The b value is the square root of the number under x, which is 1 (the square root of 1).
03

Calculate the Distance of the Foci from the Center

The foci are located at a distance \(c=\sqrt{a^{2}-b^{2}}\) from the center, in the direction of the major axis. Substituting our values, we get \(c=\sqrt{6^{2}-1^{2}}=\sqrt{35}\). The foci are then located at (-4, -3+sqrt(35)) and (-4, -3-sqrt(35)).
04

Graph the Ellipse

Plot the center of the ellipse at the point (-4, -3). From there, plot the vertical major axis with length 2a=12 units and horizontal minor axis with length 2b=2 units. Then plot the two foci. The ellipse will be the set of points such that the sum of the distances from (-4, -3+sqrt(35)) and (-4, -3-sqrt(35)) to any point on the ellipse is constant, and equals 2a=12 units. Draw the ellipse to these specs.

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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 ellipse and give the location of its foci. $$9 x^{2}+16 y^{2}-18 x+64 y-71=0$$

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