Question

How do you write an equation of an ellipse with foci $\left( 0,0 \right)$,$\left( 0,8 \right)$ and a major axis of length 16?

Answer 1

Hint: Now, we know that the equation of ellipse is given by $\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1,a>b$, where h and k are center of ellipse and a is the half length of major axis. By substituting the values we will get the desired answer.

Complete step-by-step solution:
We have been given a foci of ellipse $\left( 0,0 \right)$,$\left( 0,8 \right)$ and major axis of length 16.
We have to find the equation of ellipse.
Now, we know that the standard equation of ellipse is given by $\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1,a>b$.
Now, we know that the center of the ellipse is given by
\[\begin{align}
  & \Rightarrow C=\left( \dfrac{0+0}{2},\dfrac{0+8}{2} \right) \\
 & \Rightarrow C=\left( 0,\dfrac{8}{2} \right) \\
 & \Rightarrow C=\left( 0,4 \right) \\
\end{align}\]
Now, we know that the distance between the foci is given by
$\begin{align}
  & \Rightarrow 2c=\sqrt{{{\left( 0-0 \right)}^{2}}+{{\left( 0-8 \right)}^{2}}} \\
 & \Rightarrow 2c=\sqrt{0+64} \\
 & \Rightarrow 2c=\sqrt{64} \\
 & \Rightarrow 2c=8 \\
 & \Rightarrow c=4 \\
\end{align}$
Now, we know that the length of major axis is given as 2a
So we get
$\begin{align}
  & \Rightarrow 2a=16 \\
 & \Rightarrow a=8 \\
\end{align}$
Now, we know that focus will be finding out by using the formula ${{c}^{2}}={{a}^{2}}-{{b}^{2}}$
Now, substituting the values we will get
$\begin{align}
  & \Rightarrow {{4}^{2}}={{8}^{2}}-{{b}^{2}} \\
 & \Rightarrow 16=64-{{b}^{2}} \\
 & \Rightarrow {{b}^{2}}=64-16 \\
 & \Rightarrow {{b}^{2}}=48 \\
\end{align}$
Now, substituting all values in the standard equation of ellipse we will get
$\begin{align}
  & \Rightarrow \dfrac{{{\left( x-0 \right)}^{2}}}{{{8}^{2}}}+\dfrac{{{\left( y-4 \right)}^{2}}}{48}=1 \\
 & \Rightarrow \dfrac{{{\left( x \right)}^{2}}}{64}+\dfrac{{{\left( y-4 \right)}^{2}}}{48}=1 \\
 & \Rightarrow \dfrac{{{x}^{2}}}{64}+\dfrac{{{\left( y-4 \right)}^{2}}}{48}=1 \\
\end{align}$

Hence above is the required equation of ellipse.

Note: The point to be noted is that if the center is not origin then we get a different solution. In this particular question only the Y-coordinate changes; it means we get the ellipse with a vertical major axis. To solve such types of questions students must have an idea about the equation of the ellipse.