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25 votes
25 votes
Find the vertices and foci of the ellipse.

User Jps
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1 Answer

15 votes
15 votes

Solution

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


(x^2)/(9)+(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} ((x-h)^2)/(a^2)+((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} \\ (x^2)/(9)+(y^2)/(5)=1 \\ \\ ((x-h)^2)/(a^2)+((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}

User Ruseel
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3.4k points