Intermediate Algebra: Functions and Graphs

The graph is an ellipse with major axis on the \(y\)-axis. Because \(a^2 = 8\) and \(b^2 = 25\text{,}\) the vertices are located at \((0, 5)\) and \((0, -5)\text{,}\) and the covertices lie \(\sqrt{8} \) units to the right and left of the center, or approximately at \((2.8, 0)\) and \((-2.8, 0)\text{.}\)

To sketch the ellipse, we first locate the vertices and covertices. Then we draw a smooth curve through the points. The graph of \(\dfrac{x^2}{8}+\dfrac{y^2}{25}=1 \) is shown below.

1. We substitute \(y = \alert{2}\) into the equation and solve for \(x\text{.}\)

   \begin{align*} 4x^2+(\alert{2})^2\amp = 12\\ 4x^2\amp = 8\\ x^2\amp = 2\\ x\amp= \pm \sqrt{2} \end{align*}

   

   There are two points with \(y=2\text{,}\) namely \(\left(\sqrt{2},2\right) \) and \(\left(-\sqrt{2},2\right) \)

2. We substitute \(y = \alert{-4}\) into the equation and solve for \(x\text{.}\)

   \begin{align*} 4x^2+(\alert{-4})^2\amp = 12\\ 4x^2\amp = -4\\ x^2\amp = -1 \end{align*}

   Because there are no real solutions, there are no points on the ellipse with \(= -4\text{.}\)

1. The graph is an ellipse with center at \((-2,1)\text{.}\) We have \(a=4\) and \(b=\sqrt{5}\text{,}\) and the major axis is parallel to the \(x\)-axis because \(a \gt b\text{.}\) We plot the vertices four units to the left and right of the center, a \((-6,1)\) and \((2,1)\text{.}\) The covertices lie \(\sqrt{5}\) units above and below the center, at approximately \((-2, 3.2)\) and \((-2, -1.2)\text{.}\) The graph is shown below.

   

2. We set \(y=\alert{0}\) and solve the resulting equation to find the \(x\)-intercepts.

   \begin{align*} \dfrac{(x+2)^2}{16} + \dfrac{(\alert{0}-1)^2}{5} \amp = 1 \amp\amp \blert{\text{Subtract}~ \dfrac{1}{5}~ \text{from both sides.}}\\ \dfrac{(x+2)^2}{16} \amp = \dfrac{4}{5} \amp\amp \blert{\text{Multiply both sides by 16.}}\\ (x+2)^2 \amp = \dfrac{64}{5} \amp\amp \blert{\text{Extract roots.}}\\ x+2 \amp = \pm \sqrt{\dfrac{64}{5}}\\ x \amp = -2 \pm \dfrac{8\sqrt{5}}{5} \end{align*}

   The \(x\)-intercepts are \(\left(-2 \pm \dfrac{8\sqrt{5}}{5}, 0\right)\) or approximately \((1.6,0)\) and \((-5.6,0)\text{.}\) We set \(x=\alert{0}\) to find the \(y\)-intercepts.

   \begin{align*} \dfrac{(\alert{0}+2)^2}{16} + \dfrac{(y-1)^2}{5} \amp = 1 \amp\amp \blert{\text{Subtract}~ \dfrac{1}{4}~ \text{from both sides.}}\\ \dfrac{(y-1)^2}{5} \amp = \dfrac{3}{4} \amp\amp \blert{\text{Multiply both sides by 5.}}\\ (y-1)^2 \amp = \dfrac{15}{4} \amp\amp \blert{\text{Extract roots.}}\\ y-1 \amp = \pm \sqrt{\dfrac{15}{4}}\\ y \amp = 1 \pm \dfrac{\sqrt{15}}{2} \end{align*}

   The \(y\)-intercepts are \(\left(0, 1 \pm \dfrac{\sqrt{15}}{4}\right)\) or approximately \((0, 2.9)\) and \((0, -0.9)\)

1. We first prepare to complete the square in both \(x\) and \(y\text{.}\) Begin by factoring out the coefficients of \(x^2\) and \(y^2\text{.}\)

   \begin{gather*} \alert{4}(x^2-4x \underline{\hphantom{0000}}) + \alert{9}(y^2 - 2y \underline{\hphantom{0000}}) = 11 \end{gather*}

   We complete the square in \(x\) by adding \(\alert{4}\) to \(x^2-4x\text{,}\) and adding \(4 \cdot \alert{4}\text{,}\) or \(\blert{16}\text{,}\) to the right side of the equation. We complete the square in \(y\) by adding \(\alert{1}\) to \(y^2-2y\text{,}\) and \(9 \cdot \alert{1}\text{,}\) or \(\blert{9}\text{,}\) to the right side.

   \begin{gather*} 4(x^2-4x+\alert{4}) + 9(y^2 - 2y + \alert{1}) = 11 + \blert{16} + \blert{9} \end{gather*}

   We write each term on the left side as a perfect square to get

   \begin{align*} 4(x-2)^2 + 9(y-1)^2 \amp= 36 \amp\amp \blert{\text{Divide both sides by 36.}} \\ \frac{(x-2)^2}{9} + \frac{(y-1)^2}{4} = 1 \end{align*}

2. The graph is an ellipse with center at \((2, 1),~a^2=9\text{,}\) and \(b^2=4\) The vertices lie 3 units to the right and left of the center at \((5,1)\) and \((-1,1)\text{;}\) the covertices lie 2 units above and below the center at \((2,3)\) and \((2,-1)\text{.}\) The graph is shown below.

   

You may find it helpful to plot the given points to help you visualize the ellipse. The center of the ellipse is the midpoint of the major (or minor) axis.

\begin{align*} h \amp = \overline{x} = \dfrac{1+5}{2} = 3\\ k \amp = \overline{y} = \dfrac{3-5}{2} = -1 \end{align*}

Thus, the center is the point \((3, -1)\text{.}\) The horizontal axis is shorter, and \(a\) is the distance between the center and either covertex, say \((5, −1)\) Thus,

The value of \(b\) is the distance from the center to one of the vertices, say \((3,3)\text{:}\)

The equation of the ellipse has the form

The equation of the ellipse has \(h=3\text{,}\) \(k=-1\text{,}\) \(a=2\text{,}\) \(b=4\text{.}\) Thus the equation is

If we clear this equation of fractions and expand the powers, we obtain the general form