12.1 The ellipse  (Page 5/16)

First, we determine the position of the major axis. Because $25>9,$ the major axis is on the y -axis. Therefore, the equation is in the form $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1,$ where $b^2=9$ and $a^2=25.$ It follows that:

• the center of the ellipse is $\left(0,0\right)$
• the coordinates of the vertices are $\left(\mathrm{0,}\pm a\right)=\left(\mathrm{0,}\pm \sqrt{25}\right)=\left(\mathrm{0,}\pm 5\right)$
• the coordinates of the co-vertices are $\left(\pm b,0\right)=\left(\pm \sqrt{9},0\right)=\left(\pm 3,0\right)$
• the coordinates of the foci are $\left(\mathrm{0,}\pm c\right),$ where $c^2=a^2-b^2$ Solving for $c,$ we have:

Therefore, the coordinates of the foci are $\left(\mathrm{0,}\pm 4\right).$

Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse. See [link] .

First, use algebra to rewrite the equation in standard form.

Next, we determine the position of the major axis. Because $25>4,$ the major axis is on the x -axis. Therefore, the equation is in the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ where $a^2=25$ and $b^2=4.$ It follows that:

• the center of the ellipse is $\left(0,0\right)$
• the coordinates of the vertices are $\left(\pm a,0\right)=\left(\pm \sqrt{25},0\right)=\left(\pm 5,0\right)$
• the coordinates of the co-vertices are $\left(\mathrm{0,}\pm b\right)=\left(\mathrm{0,}\pm \sqrt{4}\right)=\left(\mathrm{0,}\pm 2\right)$
• the coordinates of the foci are $\left(\pm c,0\right),$ where $c^2=a^2-b^2.$ Solving for $c,$ we have:

Therefore the coordinates of the foci are $\left(\pm \sqrt{21},0\right).$

Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.