Chapter 10.4, Problem 48PPSE

Solution

To find: The equation of an ellipse in standard form with center at the origin, focus $\left(1,0\right)$ and width $4$ .

The equation the requires ellipse is $\frac{x^2}{5}+\frac{y^2}{4}=1$ .

Given information: The focus of the ellipse is $\left(1,0\right)$ and width is $4$ .

Formula used: The equation of the ellipse with vertices $\left(\pm a,0\right)$ and co-vertices $\left(0,\pm b\right)$ is,

  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ; $a>b>0$ .

The coordinates of the foci are $\left(\pm c,0\right)$ .

Here, $c^2=a^2-b^2$ .

Calculation:

Since, focus of the ellipse is $\left(1,0\right)$ . That means $c=1$ .

Since, the width of the ellipse is $4$

This implies that,

  $\begin{array}{c}2b=4\\ b=2\end{array}$

Now,

  $\begin{array}{c}{c}^2=a^2-b^2\\ 1^2=a^2-2^2\\ 1+2^2=a^2\\ a^2=1+4\end{array}$

Further solve,

  $\begin{array}{c}{a}^2=5\\ a=\pm \sqrt{5}\end{array}$

Then equation of the ellipse is,

  $\begin{array}{c}\frac{x^2}{{\left(\sqrt{5}\right)}^2}+\frac{y^2}{2^2}=1\\ \frac{x^2}{5}+\frac{y^2}{4}=1\end{array}$

Therefore, the required equation of the ellipse $\frac{x^2}{5}+\frac{y^2}{4}=1$ .