Chapter 8.2, Problem 6E

Solution

To calculate: The vertices and foci of the ellipse.

The vertices of the ellipse are $\left(0,3\right)$ , $\left(0,-3\right)$ and the foci are $\left(0,\sqrt{5}\right),\left(0,-\sqrt{5}\right)$ .

Given information:

The equation of ellipse is $9x^2+4y^2=36$ .

Calculation:

The vertices are the two points on the ellipse that intersect the major axis. For an ellipse with major axis parallel to the $y-$ axis, the vertices are $\left(h,k+b\right),\left(h,k-b\right)$ .

  $9x^2+4y^2=36$

Divide both sides by $12$ :

  $\begin{array}{l}\frac{9x^2}{36}+\frac{4y^2}{36}=\frac{36}{36}\\ \frac{x^2}{4}+\frac{y^2}{9}=1\end{array}$

Compare the equation with standard equation of ellipse:

Ellipse with center $\left(h,k\right)=\left(0,0\right)$ .

  $\begin{array}{l}a=\sqrt{4}=2\\ b=\sqrt{9}=3\end{array}$

Thus, the vertices of the ellipse are $\left(0,3\right)$ , $\left(0,-3\right)$ .

Calculate the value of $c:$

  $\begin{array}{l}c=\sqrt{{\left(3\right)}^2-{\left(2\right)}^2}\\ c=\sqrt{9-4}\\ c=\sqrt{5}\end{array}$

The foci of the ellipse are defined as $\left(h,k+c\right),\left(h,k-c\right)$ .

Thus, the foci are $\left(0,\sqrt{5}\right),\left(0,-\sqrt{5}\right)$ .