Question

Writing an equation of an eclipse given the foci and major axis length

Answer 1

The equation of the ellipse is:

`\begin{equation*} ((x-3)^2)/(64)+((y+3)^2)/(55)=1 \end{equation*}`

Explanation:

Remember that the general equation of an ellipse is:

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

Where:

• (h,k), are the coordinates of the center

,

• a, is the major axis lenght

,

• b, is the minor axis lenght

We also have that the equation for the foci is:

`F(h\pm c,k)`

And we also have that:

`c^2=a^2-b^2`

Since we have two foci, we can find the values of h and c as following:

`\begin{gathered} F(h\pm c,k) \\ \\ (6,-3)\rightarrow h+c=6 \\ (0,-3)\rightarrow h-c=0 \end{gathered}`

We'll have the following system of equations:

`\begin{cases}h+c=6 \\ h-c={\text{ }0}\end{cases}`

Adding up both equations and solving for h,

`\begin{cases}h+c=6 \\ h-c={\text{ }0}\end{cases}\rightarrow2h=6\rightarrow h=3`

Using this h value in the first equation and solving for c,

`h+c=6\rightarrow3+c=6\rightarrow c=3`

We'll have that the solution to this particular system of equations is:

`\begin{gathered} h=3 \\ c=3 \end{gathered}`

Now we know the value of c we can calculate the value of b as following:

`\begin{gathered} c^2=a^2-b^2 \\ \rightarrow b^=√(a^2-c^2) \\ \rightarrow b=√(8^2-3^2) \\ \\ \Rightarrow b=√(55) \end{gathered}`

With all these calculations, we'll have all the values we need to put together the equation:

`\begin{gathered} h=3 \\ k=-3 \\ a=8 \\ b=√(55) \end{gathered}`

This way, we'll have that the equation of the ellipse is:

`\begin{gathered} ((x-3)^2)/(8^2)+((y-(-3))^2)/((√(55))^2)=1 \\ \\ \Rightarrow((x-3)^2)/(64)+((y+3)^2)/(55)=1 \end{gathered}`