Chapter 10.3, Problem 17E

To determine

To Find: The standard form of equation of ellipse with the given characteristics.

Answer to Problem 17E

The equation of the ellipse is given by $\frac{x^2}{36}+\frac{5y^2}{9}=1$

Explanation of Solution

Given information:

  $Vertices=\left(\pm 6,0\right)$and passes through the point $\left(4,1\right)$

Concept and Formula Used:

The standard form of an ellipse with center $\left(h,k\right)$ and major axis parallel to x-axis is given by $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2},a>b$ with the coordinates of vertices $\left(h\pm a,k\right)$

Calculation:

The given information is that the ellipse has $vertices=\left(\pm 6,0\right)$and passes through the point $\left(4,1\right)$ , so we have the ellipse with major axis parallel to x-axis. Thus, we get

  $\begin{array}{l}\left(h\pm a,k\right)=\left(\pm 6,0\right)\\ \Rightarrow \left(h,k\right)=\left(0,0\right),a=6\end{array}$ .

Also, as it passes through $\left(4,1\right)$ , we have

  $\begin{array}{l}\frac{{\left(4\right)}^2}{{\left(6\right)}^2}+\frac{{\left(1\right)}^2}{b^2}=1\\ \frac{16}{36}+\frac{1}{b^2}=1\\ \frac{1}{b^2}=1-\frac{16}{36}\\ \frac{1}{b^2}=\frac{36-16}{36}\\ \frac{1}{b^2}=\frac{20}{36}\\ b^2=\frac{36}{20}\\ b=\frac{6}{2\sqrt{5}}=\frac{3}{\sqrt{5}}\end{array}$

So, the standard equation is given by

  $\frac{x^2}{36}+\frac{5y^2}{9}=1$

Conclusion:

The equation of ellipse is $\frac{x^2}{36}+\frac{5y^2}{9}=1$ .