Chapter 9.4, Problem 18WE

To determine

To calculate: To write the equation of ellipse with given intercepts

Answer to Problem 18WE

Equation of ellipse is $\frac{x^2}{6}+\frac{y^2}{12}=1$

Explanation of Solution

Given information: Intercepts of ellipse are $x-intercept:\pm \sqrt{6};y-intercept:\pm 2\sqrt{3}$

Formula Used:

Standard form of the equation of ellipse with center $\left(h,k\right)$ and major axis parallel to x-axis is:

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

Where, $a=$ x-intercept, $b=$ y-intercept, $c^2=a^2-b^2$

When, $a>b$ , foci lies on x-axis and coordinates of foci are $\left(h\pm c,k\right)$

When $b>a$ , foci lies on y-axis and coordinates of foci are $\left(h,k\pm c\right)$

Calculation:

Given the intercepts of ellipse as follows:

  $\begin{array}{l}x-intercept:\pm \sqrt{6}\\ y-intercept:\pm 2\sqrt{3}\end{array}$

Thus,

  $\begin{array}{l}a=\pm \sqrt{6}\\ b=\pm 2\sqrt{3}\end{array}$

Equation of ellipse is given as follows:

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

Substituting the values:

  $\begin{array}{l}\frac{{\left(x-0\right)}^2}{{\left(\pm \sqrt{6}\right)}^2}+\frac{{\left(y-0\right)}^2}{{\left(\pm 2\sqrt{3}\right)}^2}=1\\ \frac{x^2}{6}+\frac{y^2}{12}=1\end{array}$

Conclusion:

Hence, equation of ellipse is $\frac{x^2}{6}+\frac{y^2}{12}=1$