Chapter 8, Problem 43RE

Solution

To find: the equation for the conic in the standard form.

The standard equation of the ellipse with center $\left(0,2\right)$ and focal axis $y=2$ is $\frac{x^2}{9}+\frac{{\left(y-2\right)}^2}{5}=1$ .

Given data:

Type of conic: Ellipse

Center of the ellipse $=\left(0,2\right)$

Semi-major axis $=3$

One focus is $\left(2,2\right)$ .

Formula used:

The standard equation of the ellipse with center $\left(h,k\right)$ and focal axis $y=k$ is given as:

  $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$

Calculation:

It is given that the center of the ellipse is $\left(0,2\right)$ and semi-major axis is $3$ .

  $h=0,k=2,a=3$

Foci of the ellipse is given as $\left(h\pm c,k\right)$ . One of the focus is $\left(2,2\right)$ .

  $\begin{array}{c}h\pm c=2\\ 0\pm c=2\\ c=\pm 2\end{array}$

Calculate $b$ using Pythagorean theorem.

  $\begin{array}{c}{a}^2=b^2+c^2\\ 3^2=b^2+2^2\\ b^2=\left(3-2\right)\left(3+2\right)\\ b^2=5\\ b=\pm \sqrt{5}\end{array}$

The standard equation of the ellipse with center $\left(0,2\right)$ and focal axis $y=2$ is given as:

  $\begin{array}{c}\frac{{\left(x-0\right)}^2}{3^2}+\frac{{\left(y-2\right)}^2}{{\left(\sqrt{5}\right)}^2}=1\\ \frac{x^2}{9}+\frac{{\left(y-2\right)}^2}{5}=1\end{array}$

Therefore, the standard equation of the ellipse with center $\left(0,2\right)$ and focal axis $y=2$ is $\frac{x^2}{9}+\frac{{\left(y-2\right)}^2}{5}=1$ .