Problem 48 - College Algebra

Question

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

Step-by-Step Solution

Answer

The foci of the given ellipse are located at (-1, 3+√3) and (-1, 3-√3).

Step 1: Identify the Center

The given equation is \(\frac{(x+1)^{2}}{2}+\frac{(y-3)^{2}}{5}=1\). Compare it with the standard form of the ellipse equation, \((x-h)^{2}/a^{2}+(y-k)^{2}/b^{2}=1\). Here, the center (h, k) of the ellipse is at (-1, 3).

Step 2: Identify the Lengths of Semi-Major and Semi-Minor axis

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

Step 3: Calculate the Distance from Center to each Focus

The distance from the center to each focus or the focal length (c) is given by \(c=\sqrt{|a^{2}-b^{2}|\}\). Put the values of a and b in the formula, we get \(c=\sqrt{5-2} = \sqrt{3}\).

Step 4: Find the Foci

Since, the ellipse is vertical (\(a^{2}\) is under \(y^{2}\)), the foci are located at \(k \pm c\), on the major axis. So the foci of the ellipse are at (-1, 3+√3) and (-1, 3-√3).

Step 5: Draw the Ellipse

Now we can depict the ellipse on the coordinate plane. Plot the center at (-1,3), the vertices at (-1, 3+√5) and (-1, 3-√5) and the foci as found in the previous step. Sketch the ellipse through these points.