Question

Find the standard form of the equation of the ellipse that satisfies the given conditions:Vertices: (-2, -1) and (8, -1)Foci: (-1, -1) and (7, -1)

Answer 1

The standard equation of an ellipse is expressed as

`\begin{gathered} ((x-h)^2)/(a)+((y-k)^2)/(b)=1\text{ } \\ \text{where} \\ (h,k)\text{ are the coordinate of the center of the ellipse.} \end{gathered}`

where

`\begin{gathered} \text{the vertices}\Rightarrow(h\pm a,k),(h,k\pm b)_{} \\ \text{foci}\Rightarrow(h\pm c,k) \\ \text{where} \\ c^2=a^2-b^2 \end{gathered}`

Step 1:

Evaluate the center (h,k) along the vertices (-2,-1) and (8, -1)

Using the midpoint formula

`\begin{gathered} (h,k)\text{ = (}(-2+8)/(2),\text{ }(-1+(-1))/(2)) \\ =((6)/(2),-(2)/(2)) \\ \Rightarrow(h,k)=(3,\text{ -1)} \end{gathered}`

Step 2:

From the length of the major axis of the ellipse given as 2a, we have

`\begin{gathered} 2a=\sqrt[]{(}8-(-2))^2-(-1-(-1))^2 \\ 2a=10 \\ \Rightarrow a=5 \end{gathered}`

Step 3:

Recall that the foci is expressed as

`\begin{gathered} \text{foci: }(h\pm c,k) \\ \text{thus, } \\ h+c=7,\text{ k=-1 or h-c=-1, k=-1} \\ where\text{ h=3} \\ \Rightarrow c=7-3=4 \end{gathered}`

but

`\begin{gathered} c^2=a^2-b^2 \\ \Rightarrow16=25-b^2 \\ b^2=25-16=9 \\ \Rightarrow b=3 \end{gathered}`

Hence, the standard form of the ellipse becomes

`((x-3)^2)/(5)+((y+1)^2)/(3)=1`