Chapter 8.2, Problem 5E

Solution

To calculate: The vertices and foci of the ellipse.

The vertices of the ellipse are $\left(2,0\right)$ , $\left(-2,0\right)$ and the foci are $\left(1,0\right),\left(-1,0\right)$ .

Given information:

The equation of ellipse is $3x^2+4y^2=12$ .

Calculation:

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

  $3x^2+4y^2=12$

Divide both sides by $12$ :

  $\begin{array}{l}\frac{3x^2}{12}+\frac{4y^2}{12}=\frac{12}{12}\\ \frac{x^2}{4}+\frac{y^2}{3}=1\end{array}$

Compare the equation with standard equation of ellipse:

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

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

Thus, the vertices of the ellipse are $\left(2,0\right)$ , $\left(-2,0\right)$ .

Calculate the value of $c:$

  $\begin{array}{l}c=\sqrt{{\left(2\right)}^2-{\left(\sqrt{3}\right)}^2}\\ c=\sqrt{4-3}\\ c=\sqrt{1}\\ c=1\end{array}$

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

Thus, the foci are $\left(1,0\right),\left(-1,0\right)$ .