Question

Can you help me with 23The directions say“ for the following exercises write the equation of an ellipse in standard form and identify the end points of the major and minor axes as well as the foci

Answer 1

In general, the standard form of an ellipse centered at (h,k) is

`((x-h)^2)/(a^2)+((y-k)^2)/(b^2)=1`

In our case, given

`4x^2+40x+25y^2-100y+100=0`

Complete squares as shown below

`\begin{gathered} 4x^2+40x=(2x)^2+2(2x)p \\ \Rightarrow40x=4px \\ \Rightarrow p=10 \\ \Rightarrow4x^2+40x+100=(2x+10)^2 \end{gathered}`

Similarly,

`\begin{gathered} 25y^2-100y=(5y)^2+2\cdot q\cdot5y \\ \Rightarrow-100=10q \\ \Rightarrow q=-10 \\ \Rightarrow25y^2-100y+100=(5y-10)^2 \end{gathered}`

Thus, the initial expression can be transform in the following way,

`\begin{gathered} \Rightarrow4x^2+40x+25y^2-100y+100+100=0+100 \\ \Rightarrow(2x+10)^2+(5y-10)^2=100 \end{gathered}`

Therefore,

`\begin{gathered} \Rightarrow2^2(x+5)^2+5^2(y-2)^2=100 \\ \Rightarrow(2^2(x+5)^2+5^2(y-2)^2)/(100)=1 \\ \Rightarrow((x+5)^2)/(25)+((y-2)^2)/(4)=1 \\ \Rightarrow((x+5)^2)/(5^2)+((y-2)^2)/(2^2)=1 \end{gathered}`

This last line is the standard form of the ellipse.

Notice that it is centered at (-5,2), the major axis is equal to 5 and the minor one is equal to 2. Therefore, the endpoints are

`\begin{gathered} (-5+5,2)=(0,2) \\ (-5-5,2)=(-10,2) \\ (-5,2-2)=(-5,0) \\ (-5,2+2)=(-5,4) \end{gathered}`

Finally, as for the foci, we can find them given the major and minor axis using the formula below

`\begin{gathered} c=\sqrt[]{a^2-b^2} \\ (\pm c-5,2)\to foci \\ a\to\text{major axis} \\ b\to\text{ minor axis} \end{gathered}`
`\begin{gathered} \Rightarrow c=\sqrt[]{25-4}=\sqrt[]{21} \\ \Rightarrow(-\sqrt[]{21}-5,2),(\sqrt[]{21}-5,2) \end{gathered}`

Foci: (-sqrt21-5,2), (sqrt21-5,2)