Chapter 9, Problem 28RE

To determine

To find: The centre, vertices, foci, and eccentricity of the ellipse $\frac{y^2}{49}+\frac{x^2}{2}=1$ , and sketch it graph.

Answer to Problem 28RE

Center: $\left(0,0\right)$ .

Vertices: $\left(0,-7\right)$ and $\left(0,7\right)$ ,

Foci: $\left(0,-\sqrt{47}\right)$ and $\left(0,\sqrt{47}\right)$

Eccentricity: $\frac{\sqrt{47}}{7}$

Explanation of Solution

Given:

  $\frac{y^2}{49}+\frac{x^2}{2}=1$

Calculation:

The standard equation of an ellipse is given by $\frac{{\left(y-k\right)}^2}{a^2}+\frac{{\left(x-h\right)}^2}{b^2}=1$ .

The center of this ellipse is $\left(h,k\right)$ .

Write the given ellipse in standard form.

  $\frac{{\left(y-0\right)}^2}{7^2}+\frac{{\left(x-0\right)}^2}{{\left(\sqrt{2}\right)}^2}=1$

Compare the given ellipse $\frac{{\left(y-0\right)}^2}{7^2}+\frac{{\left(x-0\right)}^2}{{\left(\sqrt{2}\right)}^2}=1$ with the standard form $\frac{{\left(y-k\right)}^2}{a^2}+\frac{{\left(x-h\right)}^2}{b^2}=1$ . This gives,

  $h=0$ , $k=0$ , $a=7$ and $b=\sqrt{2}$ .

So, the center of the ellipse is $\left(0,0\right)$ .

Since $b>a$ , the ellipse is vertical. So, the vertices of the ellipse are $\left(0,-7\right)$ and $\left(0,7\right)$ .

Now, find the value of $c$ using the equation $b^2-c^2=a^2$ .

  $\begin{array}{l}{a}^2-c^2=b^2\\ 7^2-c^2={\left(\sqrt{2}\right)}^2\\ \Rightarrow c=\sqrt{49-2}=\sqrt{47}\end{array}$

So, the foci of the ellipse are $\left(0,-\sqrt{47}\right)$ and $\left(0,\sqrt{47}\right)$ .

Then, the value of the eccentricity is $e=\frac{c}{a}=\frac{\sqrt{47}}{7}$ .

Now, graph the ellipse as shown below.

  

Now, label the center, vertices and foci on the graph to verify the result as shown below.