Chapter 11, Problem 15RE

To determine

To find: The center, vertices, foci and length of major and minor axis of ellipse and represent it through a graph.

Answer to Problem 15RE

Center is at $\left(0,2\right)$

Focii $=\left(\sqrt{5},2\right)\text{ and }\left(-\sqrt{5},2\right)$

Vertices $=\left(-2,2\right)\text{ and }\left(2,2\right)$

Major axis length $=6$ , minor axis length $=4$

Explanation of Solution

Given information:

Given equation of the ellipse is $4x^2+9y^2=36y$

Calculation:

If the equation of the ellipse:-

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

Then center of the ellipse is $\left(h,k\right)$

If $a>b$

Then length of the major axis $=2a$

Length of the minor axis $=2b$

If $b>a$

Then length of major axis $=2b$

Then length of minor axis $=2a$

If $a>b$

Coordinate of the vertices are $\left(b\pm a,k\right)$

Coordinates of the foci are $\left(h\pm c,k\right)$

  $c^2=a^2-b^2$

If $b>a$

Coordinate of the vertices $\Rightarrow \left(h,k\pm a\right)$

Coordinate of the foci $\Rightarrow \left(h,k\pm c\right)$

  $c^2=a^2+b^2$

Convert $4x^2+9y^2=3by$ to

  $\begin{array}{l}\frac{{\left(x-a\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1\\ 4x^2+9y^2=36y\\ \Rightarrow 4x^2+9yh2-36y=0\\ 4x^2+9y^2-36y+36-36=0\\ {\left(2x\right)}^2+{\left(3y-6\right)}^2=26\\ \frac{x^2}{9}+\frac{{\left(y-2\right)}^2}{4}=1\\ \Rightarrow \text{Now }h=0,k=2\\ a=3,b=2\end{array}$

  $a>b$

Thus, the length of the major axis $=2a=2\left(3\right)=6$

Length of the minor axis $=2b=2\times 2=4$

Cenetr of the ellipse $=\left(0,2\right)$

Coordinate of vertices $=\left(h\pm a,k\right)$

  $=\left(\pm 2,2\right)\Rightarrow \left(2,2\right)\text{ and }\left(-2,2\right)$

Coordinate of foci $\Rightarrow \left(h\pm c,k\right)$

  $\begin{array}{l}{c}^2=a^2-b^2\\ c^2=5\Rightarrow c=\sqrt{5}\end{array}$

Foci $\Rightarrow \left(\sqrt{5},2\right)\text{ and }\left(-\sqrt{5},2\right)$

  

Interpretation :

Through the equation of the ellipse the following values are computed

Center is at $\left(0,2\right)$

Focii $=\left(\sqrt{5},2\right)\text{ and }\left(-\sqrt{5},2\right)$

Vertices $=\left(-2,2\right)\text{ and }\left(2,2\right)$

Major axis length $=6$ , minor axis length $=4$