Question

Find the vertices and foci of the ellipse.

Answer 1

Solution

- The question would like us to find the vertices and foci of the ellipse with the following equation:

$\frac{x^2}{9}+\frac{y^2}{5}=1$

- Before we can solve, we need to write down the general equation of an ellipse and formulas for the foci and vertices.

- These are given below:

$\begin{gathered} \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1 \\ \text{where,} \\ \text{Vertices are:} \\ (h+a,k),(h-a,k),(h,k+b),(h,k-b) \\ \\ \text{The foci are:} \\ (\pm c,0),\text{ where, }c^2=a^2-b^2\text{ (If the major axis is on the x-axis)} \\ \\ (\text{if }a>b\text{, then, the major axis is on the x)} \end{gathered}$

- Based on the formulas given above, we can proceed to solve as follows:

$\begin{gathered} \text{ Compare the general equation of an ellipse to the equation given to us} \\ \frac{x^2}{9}+\frac{y^2}{5}=1 \\ \\ \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1 \\ \\ h=k=0 \\ a^2=9,b^2=5 \\ a=\pm3,b=\pm\sqrt[]{5} \\ \\ \text{ We can s}ee\text{ that } \\ a>b\implies3>\sqrt[]{5} \\ \text{Thus, the major axis is on the x} \\ \\ \text{Thus, the Vertices are:} \\ (h+a,k)=(0+3,0)=(3,0) \\ (h-a,k)=(0-3,0)=(-3,0) \\ (h,k+b)=(0,0+\sqrt[]{5})=(0,\sqrt[]{5)} \\ (h,k-b)=(0,0-\sqrt[]{5})=(0,-\sqrt[]{5}) \\ \\ \text{The foci is gotten as follows:} \\ a^2-b^2=c^2 \\ 9-5=c^2 \\ c^2=4 \\ c=\pm\sqrt[]{4} \\ c=\pm2 \\ \text{Thus, the foci are:} \\ (\pm2,0) \end{gathered}$

Final Answer

The answers are:

$\begin{gathered} \text{Vertices are:} \\ (3,0),(-3,0),(0,\sqrt[]{5}),(0,-\sqrt[]{5}) \\ \\ \text{Foci are:} \\ (2,0),(-2,0) \end{gathered}$