Chapter 11.1, Problem 61PPSE

Solution

To identify: The center, vertices, and foci for the ellipse $\frac{{\left(x-2\right)}^2}{9}+\frac{{\left(y-1\right)}^2}{25}=1$ .

The center is $\left(h,k\right)=\left(2,1\right)$ , the vertices are $\left(2,6\right)\text{and}\left(2,-4\right)$ , and the foci are $\left(2,5\right)\text{ and }\left(2,-3\right)$ .

Given information:

The given ellipse is $\frac{{\left(x-2\right)}^2}{9}+\frac{{\left(y-1\right)}^2}{25}=1$ .

Explanation:

Consider the ellipse: $\frac{{\left(x-2\right)}^2}{9}+\frac{{\left(y-1\right)}^2}{25}=1$

The standard equation form of ellipse is: $\frac{{\left(x-h\right)}^2}{b^2}+\frac{{\left(y-k\right)}^2}{a^2}=1$

The center of the given ellipse is $\left(h,k\right)=\left(2,1\right)$ .

The value of $b^2$ is 9 then $b=3$ . Similarly, the value of $a^2$ is 25 then $a=5$ .

To find the vertices of the given ellipse, use the formula: $\left(h,k\pm 1\right)$

Then the vertices are:

  $\begin{array}{c}\left(h,k-a\right)=\left(2,1-5\right)\\ =\left(2,-4\right)\\ \left(h,k+a\right)=\left(2,1+5\right)\\ =\left(2,6\right)\end{array}$

Therefore, the vertices are $\left(2,6\right)\text{and}\left(2,-4\right)$ .

Now find the foci. First find the value of c then use the formula for foci $\left(h,k\pm c\right)$ .

  $\begin{array}{l}{c}^2=b^2-a^2\\ c^2=25-9\\ c^2=16\\ c=4\end{array}$

Now the foci are:

  $\begin{array}{c}\left(h,k+c\right)=\left(2,1+4\right)\\ =\left(2,5\right)\\ \left(h,k-c\right)=\left(2,1-4\right)\\ =\left(2,-3\right)\end{array}$

Therefore, the foci is $\left(2,5\right)\text{ and }\left(2,-3\right)$ .