Chapter 9.4, Problem 14WE

To determine

To calculate: To graph the ellipse and find the foci

Answer to Problem 14WE

Foci are $\left(0,\pm \frac{\sqrt{5}}{2}\right)$

Explanation of Solution

Given information: Ellipse is $9x^2+4y^2=9$

Formula Used:

Standard form of the equation of ellipse with center $\left(h,k\right)$ and major axis parallel to x-axis is:

  $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$

Where, $a=$ x-intercept, $b=$ y-intercept, $c^2=a^2-b^2$

When, $a>b$ , foci lies on x-axis and coordinates of foci are $\left(h\pm c,k\right)$

When $b>a$ , foci lies on y-axis and coordinates of foci are $\left(h,k\pm c\right)$

Calculation:

Equation of ellipse is given as follows:

  $9x^2+4y^2=9$

Divide both sides of equation by $9$ :

  $\begin{array}{l}\frac{x^2}{1}+\frac{y^2}{\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}=1\\ \frac{x^2}{1^2}+\frac{y^2}{{\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^2}=1\end{array}$

Comparing the equation with standard form of the equation of ellipse $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$ :

  $\begin{array}{l}a=1\\ b=\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ \left(h,k\right)=\left(0,0\right)\end{array}$

Since $b>a$ , thus foci lies on y-axis.

Also, foci is calculated as follows:

  $c^2=b^2-a^2$

Substituting the values:

  $\begin{array}{l}{c}^2={\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^2-1^2\\ c^2=\frac{9}{4}-1\\ c^2=\frac{5}{4}\\ c=\pm \frac{\sqrt{5}}{2}\end{array}$

Thus, foci are $\left(0,\pm \frac{\sqrt{5}}{2}\right)$

Plotting the graph of ellipse:

  

Conclusion:

Hence, foci are $\left(0,\pm \frac{\sqrt{5}}{2}\right)$