Chapter 9.2, Problem 60E

To determine

Todraw: Thegraph of the ellipse $\frac{x^2}{4}+\frac{y^2}{25}=1$ using latera recta.

Answer to Problem 60E

The graph of the ellipse $\frac{x^2}{4}+\frac{y^2}{25}=1$ is shown below.

Explanation of Solution

Given information:

The equation of ellipse is $\frac{x^2}{4}+\frac{y^2}{25}=1$ .

Calculation:

Compare the given equation of ellipse with standard equation $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$ .

  $h=0,k=0,a=2,b=5$

Calculate the foci and latus rectum of the ellipse.

  $\begin{array}{l}c=\pm \sqrt{b^2-a^2}\\ =\pm \sqrt{25-4}\\ =\pm \sqrt{21}\end{array}$

  $\begin{array}{l}\text{Foci}=\left(0,c\right)\\ =\left(0,\pm \sqrt{21}\right)\end{array}$

Calculate the length of the latus rectum.

  $\begin{array}{l}\text{Length of latus rectum}=\frac{2a^2}{b}\\ =\frac{2\times 4}{5}\\ =\frac{8}{5}\end{array}$

Plot the graph of ellipse with the calculated data using a graphing utility.