Chapter 8.2, Problem 37E

Solution

To Find: The center, vertices, and foci of the ellipse.

Center of the ellipse $\left(-1,2\right)$ , vertices of the ellipse are $\left(4,2\right),\left(-6,2\right)$ and the foci are $\left(2,2\right),\left(-4,2\right)$ .

Given information:

The equation of ellipse is $\frac{{\left(x+1\right)}^2}{25}+\frac{{\left(y-2\right)}^2}{16}=1$ .

Calculation:

The standard form of the ellipse equation is $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$ with the center $\left(h,k\right)$ and $a,b$ are the semi-major and semi-minor axes.

Rewrite $\frac{{\left(x+1\right)}^2}{25}+\frac{{\left(y-2\right)}^2}{16}=1$ in the standard ellipse equation, it follows:

  $\frac{{\left(x-\left(-1\right)\right)}^2}{5^2}+\frac{{\left(y-2\right)}^2}{4^2}=1$

Therefore, the ellipse properties are:

  $\left(h,k\right)=\left(-1,2\right),a=5,b=4$

  $a>b$ therefore $a$ is semi-major axis $b$ is semi-minor axis.

Ellipse with center $\left(h,k\right)=\left(-1,2\right),a=5,b=4$

The vertices are the two points on the ellipse that intersect the major axis. For an ellipse with major axis parallel to the $y-$ axis, the vertices are $\left(h,k+b\right),\left(h,k-b\right)$ .

  $\frac{{\left(x+1\right)}^2}{25}+\frac{{\left(y-2\right)}^2}{16}=1$

Compare the equation with standard equation of ellipse:

Ellipse with center $\left(h,k\right)=\left(-1,2\right)$ .

  $\left(-1+5,2\right),\left(-1-5,2\right)=\left(4,2\right),\left(-6,2\right)$

Thus, the vertices of the ellipse are $\left(4,2\right),\left(-6,2\right)$ .

Calculate the value of $c:$

  $\begin{array}{l}c=\sqrt{{\left(5\right)}^2-{\left(4\right)}^2}\\ c=\sqrt{25-16}\\ c=\sqrt{9}\\ c=3\end{array}$

The foci of the ellipse are defined as $\left(h+c,k\right),\left(h-c,k\right)=\left(-1+3,2\right),\left(-1-3,2\right)$ .

Thus, the foci are $\left(2,2\right),\left(-4,2\right)$ .