Problem 47 - 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 \((1+\sqrt{3}, -3)\) and \((1-\sqrt{3}, -3)\)

Step 1: 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)

Step 2: 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 \(b=\sqrt{2}\)

Step 3: 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 \((1-\sqrt{3},-3)\)

Step 4: 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.