Chapter 9.4, Problem 1GP

To determine

To Find: The equation of ellipse with vertex at $\left(-4,0\right)\text{ and }\left(4,0\right)$ and foci at $\left(2,0\right)\text{ and }\left(-2,0\right)$ .

Answer to Problem 1GP

  $\frac{x^2}{16}+\frac{y^2}{12}=1$

Explanation of Solution

Given: Vertex at $\left(-4,0\right)\text{ and }\left(4,0\right)$ and foci at $\left(2,0\right)\text{ and }\left(-2,0\right)$ .

Equation of ellipse: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Vertex: $\left(\pm a,0\right)$

Foci: $\left(\pm c,0\right)$

The given vertex, $\left(\pm a,0\right)\iff \left(\pm 4,0\right)$

So, $a=4$

The given foci, $\left(\pm c,0\right)\iff \left(\pm 2,0\right)$

So, $c=2$

Using formula, $c^2=a^2-b^2$

  $\begin{array}{l}{2}^2=4^2-b^2\\ b^2=12\end{array}$

Now put a and b into formula.

  $\frac{x^2}{16}+\frac{y^2}{12}=1$