Chapter 10.8, Problem 36E

To determine

Write the system of equations for graph.

Answer to Problem 36E

  $\begin{array}{l}\frac{x^2}{16}+\frac{y^2}{25}\le 1\\ \frac{x^2}{9}+\frac{{\left(y+1\right)}^2}{4}>1\end{array}$

Explanation of Solution

Given information:

  

Calculation:

Consider the graph,

  

As per the graph the center of outer ellipse is $\left(0,0\right)$

Length of semi major axis $b=5$

Length of semi minor axis $a=4$

The quadratic equation for outer figure is,

  $\begin{array}{l}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ \frac{x^2}{16}+\frac{y^2}{25}=1\end{array}$

the center of inner ellipse is $\left(0,-1\right)$

Length of semi major axis $a=3$

Length of semi minor axis $b=2$

The quadratic equation for outer figure is,

  $\begin{array}{l}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ \frac{x^2}{9}+\frac{{\left(y+1\right)}^2}{4}=1\end{array}$

Let take a test point at $\left(0,0\right)$ which is inside the both figures,

First take outer figure,

  $\begin{array}{l}LHS=\frac{x^2}{16}+\frac{y^2}{25}\\ =0\\ RHS=1\\ LHS\le RHS\\ soinequalitywillbe\\ \frac{x^2}{16}+\frac{y^2}{25}\le 1\end{array}$

Now take inner figure,

  $\begin{array}{l}LHS=\frac{x^2}{9}+\frac{{\left(y+1\right)}^2}{4}\\ =\frac{0}{9}+\frac{{\left(-1+1\right)}^2}{4}=0\\ RHS=1\\ LHS\le RHS\end{array}$

The shaded region is outside the inner ellipse, so origin does not include inside it ,

The equation for outer region of the inner ellipse,

  $LHS>RHS$

The inequality will be,

  $\frac{x^2}{9}+\frac{{\left(y+1\right)}^2}{4}>1$

Hence, the systems of inequalities are,

  $\begin{array}{l}\frac{x^2}{16}+\frac{y^2}{25}\le 1\\ \frac{x^2}{9}+\frac{{\left(y+1\right)}^2}{4}>1\end{array}$