Chapter 8.6, Problem 10E

Solution

Graph the given equation and identify the important characteristics of graph.

The graph of the given equation is

  

With centre at $\left(-2,2\right)$ , and vertices $\left(0,0\right),\left(0,4\right),\left(5,2\right),\left(-5,2\right)$

Given information:

Given equation is:

  $\frac{x^2}{25}+\frac{{\left(y-2\right)}^2}{4}=1$

Formula used:

• Standard form of equation of a ellipses with center $\left(h,k\right)$ and vertices $\left(h\pm a,k\right)$ and $\left(h,k\pm b\right)$ is

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

• If $a>b$ , then ellipses has horizontal orientation and if $a<b$ , then ellipses has vertical orientation

Calculation:

The given equation $\frac{x^2}{25}+\frac{{\left(y-2\right)}^2}{4}=1$ can be rewritten as

  $\frac{{\left(x-0\right)}^2}{5^2}+\frac{{\left(y-2\right)}^2}{2^2}=1$

Now in order to graph the given equation

  $\frac{{\left(x-0\right)}^2}{5^2}+\frac{{\left(y-2\right)}^2}{2^2}=1$ ,

First identify its center and vertices by comparing the equation with standard form of the equation of the ellipses:

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

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

  $\left(h,k\right)=\left(0,2\right)$

Thus, the given ellipses has centre at $\left(-2,2\right)$

Now vertices of the ellipses are $\left(h\pm a,k\right)$ and $\left(h,k\pm b\right)$

That is, $\left(0\pm 5,2\right)=\left(5,2\right),\left(-5,2\right)$ and $\left(0,2\pm 2\right)=\left(0,0\right),\left(0,4\right)$

Also, since $a>b$ , then ellipses has horizontal orientation

Thus, the graph of the ellipses will be as

  

With centre at $\left(-2,2\right)$ , and vertices $\left(0,0\right),\left(0,4\right),\left(5,2\right),\left(-5,2\right)$