Chapter 9.4, Problem 11WE

To determine

To calculate: To graph the ellipse and find the foci

Answer to Problem 11WE

Foci are $\left(0,\pm 2\right)$

Explanation of Solution

Given information: Ellipse is $4x^2+3y^2=48$

Formula Used:

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

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

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:

  $4x^2+3y^2=48$

Divide both sides of equation by $48$ :

  $\begin{array}{l}\frac{x^2}{12}+\frac{y^2}{16}=1\\ \frac{x^2}{{\left(2\sqrt{3}\right)}^2}+\frac{y^2}{4^2}=1\end{array}$

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

  $\begin{array}{l}a=2\sqrt{3}\\ b=4\\ \left(h,k\right)=\left(0,0\right)\end{array}$

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

Also, foci is calculated as follows:

  $c^2=b^2-a^2$

Substituting the values:

  $\begin{array}{l}{c}^2=4^2-{\left(2\sqrt{3}\right)}^2\\ c^2=16-12\\ c^2=4\\ c=\pm 2\end{array}$

Thus, foci are $\left(0,\pm 2\right)$

Plotting the graph of ellipse:

  

Conclusion:

Hence, foci are $\left(0,\pm 2\right)$