Chapter 11.2, Problem 41E

To determine

To determine the equation of the ellipse.

Answer to Problem 41E

The equation of the ellipse is $\frac{{\text{x}}^2}{25}+\frac{{\text{y}}^2}{5}=1$ .

Explanation of Solution

Given information:

The given conditions are:

Length of the major axis: $10$

Foci lies on the x-axis

Ellipse passes through the point $\left(\sqrt{5},2\right)$

Formula used:

The following formula is used:

  $\text{Length of major axis}=2\text{a}$

Calculation:

The foci lies on the x-axis, so the ellipse has a major horizontal axis and ${\text{a}}^2$ is the denominator of ${\text{x}}^2$ .

The value of $\text{a}$ can be calculated by:

  $\begin{array}{l}\text{Length of major axis}=10\\ \text{2a}=10\\ \text{a}=5\end{array}$

The value of $\text{b}$ can be calculated by:

  $\begin{array}{l}\frac{{\text{x}}^2}{{\text{a}}^2}+\frac{{\text{y}}^2}{{\text{b}}^2}=1\\ \frac{{\sqrt{5}}^2}{5^2}+\frac{2^2}{{\text{b}}^2}=1\left[\text{Substituting the values of x,y and a}\right]\\ \frac{5}{25}+\frac{4}{{\text{b}}^2}=1\\ \frac{4}{{\text{b}}^2}=1-\frac{5}{25}\\ \frac{4}{{\text{b}}^2}=\frac{20}{25}\\ {\text{b}}^2=\frac{25}{20}\times 4\\ {\text{b}}^2=5\\ \text{b}=\pm \sqrt{5}\end{array}$

The equation can be obtained by:

  $\begin{array}{l}\frac{{\text{x}}^2}{{\text{a}}^2}+\frac{{\text{y}}^2}{{\text{b}}^2}=1\\ \text{or }\frac{{\text{x}}^2}{5^2}+\frac{{\text{y}}^2}{{\sqrt{5}}^2}=1\\ \text{or }\frac{{\text{x}}^2}{25}+\frac{{\text{y}}^2}{5}=1\end{array}$

Hence, the equation of the ellipse is $\frac{{\text{x}}^2}{25}+\frac{{\text{y}}^2}{5}=1$ .