College Algebra, Chapter 8, 8.4, Section 8.4, Problem 28

Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellipse, find the center, foci, vertices and lengths of the major and minor axes. If it is a parabola, find the vertex, focus and directrix. If it is a hyperbola, find the center, foci, vertices and asymptotes. Sketch the graph of the equation. If the equation has no graph, explain why.


$\begin{equation} \begin{aligned} 2x^2 + y^2 =& 2y + 1 && \text{Group terms} \\ \\ 2x^2 (y^2 - 2y) =& 1 && \text{Complete the square: add } \left( \frac{-2}{2} \right)^2 = 1 \\ \\ 2x^2 + (y^2 - 2y + 1) =& 1 + 1 && \text{Perfect square} \\ \\ 2x^2 + (y - 1)^2 =& 2 && \text{Divide both sides by } 2 \\ \\ x^2 + \frac{(y - 1)^2}{2} =& 1 && \end{aligned} \end{equation}$


The equation is an ellipse that is shifted so that its center is at $(0, 1)$. It is obtained from the ellipse $\displaystyle x^2 + \frac{y^2}{2} = 1$ with center at the origin by shifting $1$ unit upward. The endpoints of the major and minor axis of the unshifted ellipse are $(0, \sqrt{2}), (0, - \sqrt{2}), (1, 0)$ and $(-1, 0)$. By applying transformations, the corresponding end points will be..


$\begin{equation} \begin{aligned} (0, \sqrt{2}) \to (0, \sqrt{2} + 1) =& (0, \sqrt{2} + 1) \\ \\ (0, - \sqrt{2}) \to (0, - \sqrt{2} + 1) =& (0, - \sqrt{2} + 1) \\ \\ (1, 0) \to (1, 0 + 1) =& (1,1) \\ \\ (-1,0 ) \to (-1, 0 + 1) =& (-1, 1) \end{aligned} \end{equation}$


To find the foci of the shifted ellipse, we first find the foci of the unshifted ellipse. Since $a^2 = 2$ and $b^2 = 1$, then $c^2 = 2 - 1 = 1$, so $c = 1$. So the foci are $(0, \pm 1)$. By applying transformation, the foci will be

Therefore, the focus is at

$(0, 1) \to (0, 1 + 1) = (0, 2)$

$(0, -1) \to (0, -1 +1) = (0, 0)$

Therefore, the graph is