Problem 50 - College Algebra

Question

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

Step-by-Step Solution

Answer

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 1: 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).

Step 2: 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).

Step 3: 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)).

Step 4: 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.