Chapter 11.3, Problem 9MRE

To determine

To calculate: The equation of the ellipse

Answer to Problem 9MRE

The equation of the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$

Explanation of Solution

Given information: An ellipse has

  $x$ -intercepts are $\left(2,0\right)\text{ and }\left(-2,0\right)$

  $y$ -intercepts are $\left(\sqrt{3},0\right)\text{ and }\left(-\sqrt{3},0\right)$

Formula used: Distance between two points $d=\sqrt{{\left(X_2-X_1\right)}^2+{\left(Y_2-Y_1\right)}^2}$

Distance between $x$ -intercepts are $\left(2,0\right)\text{ and }\left(-2,0\right)$

  $\begin{array}{l}\therefore x=\sqrt{{\left(2+2\right)}^2+{\left(0+0\right)}^2}\\ =\sqrt{4^2}\\ =4\end{array}$

Distance between $y$ -intercepts are $\left(\sqrt{3},0\right)\text{ and }\left(-\sqrt{3},0\right)$

  $\begin{array}{l}\therefore y=\sqrt{{\left(\sqrt{3}+\sqrt{3}\right)}^2+{\left(0+0\right)}^2}\\ =\sqrt{{\left(2\sqrt{3}\right)}^2}\\ =2\sqrt{3}\end{array}$

Since, $x>y$

An ellipse is parallel to x -axis

Therefore, $2a=4\text{ and }2b=2\sqrt{3}$

Or, $a=2\text{ and }b=\sqrt{3}$

From the equation of the ellipse:

  $\therefore \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Calculation:

  $\begin{array}{l}\Rightarrow \frac{x^2}{{\left(2\right)}^2}+\frac{y^2}{{\left(\sqrt{3}\right)}^2}=1\\ \Rightarrow \frac{x^2}{4}+\frac{y^2}{3}=1\end{array}$

Hence, the equation of the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$