Chapter SH, Problem 9.12EP

To determine

To find: the equation of the ellipse shown in the graph.

Answer to Problem 9.12EP

The equation of ellipse is $\text{9}{\left(\text{x-3}\right)}^{\text{2}}\text{+25}{\left(\text{y+2}\right)}^{\text{2}}\text{=225}$ .

Explanation of Solution

Given:

  

Concept used:

The equation of the ellipse if major axis at horizontal will be:

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

Major axis $\text{=2a}$ .

Minor axis $\text{=2b}$ .

Calculation:

For ellipse $\text{a}$ is major axis and $\text{b}$ is minor axis.

From the given graph:

Center: $\left(\text{h,k}\right)=\left(3,-2\right)$ .

Major axis is at horizontal:

  $\text{2a=10}$ .

  $\text{a=5}$ .

Minor axis at vertical:

  $\text{2b=6}$ .

  $\text{b=3}$ .

So the equation of the ellipse will be:

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

  $\frac{{\left(\text{x-3}\right)}^{\text{2}}}{{\text{5}}^{\text{2}}}\text{+}\frac{{\left(\text{y-}\left(\text{-2}\right)\right)}^{\text{2}}}{{\text{3}}^{\text{2}}}\text{=1}$ .

  $\frac{{\left(\text{x-3}\right)}^{\text{2}}}{\text{25}}\text{+}\frac{{\left(\text{y+2}\right)}^{\text{2}}}{\text{9}}\text{=1}$ .

  $\frac{9{\left(\text{x-3}\right)}^{\text{2}}+\text{25}{\left(\text{y+2}\right)}^{\text{2}}}{\text{25}\times \text{9}}\text{=1}$ .

Hence

the equation of ellipse is $\text{9}{\left(\text{x-3}\right)}^{\text{2}}\text{+25}{\left(\text{y+2}\right)}^{\text{2}}\text{=225}$ .