Chapter 9.8, Problem 4P

To determine

To calculate:

The equation of an ellipse

Answer to Problem 4P

The equation of an ellipse is $\frac{x^2}{4}+\frac{y^2}{\left(\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)}=1$ .

Explanation of Solution

Given information:

  $\begin{array}{l}A=12unit{s}^2\\ P=14units\end{array}$

Formula used:

The area of rectangle is given by the following expression:

  $A=xy$

The perimeter of rectangle is given by the following expression:

  $P=2x+2y$

Where,

Length of rectangle: $x$

Width of rectangle: $y$

Calculation:

The area of rectangle is given as:

  $A=xy$

Put the given data:

  $12=xy$  ... (1)

The perimeter of rectangle is given as:

  $P=2x+2y$

Put the given data:

  $\begin{array}{l}2x+2y=14\\ x+y=7\end{array}$

  $y=7-x$  ... (2)

Now, put the value $y$ in eq. (1):

  $\begin{array}{l}12=x\left(7-x\right)\\ 12=7x-x^2\\ x^2-7x+12=0\\ \left(x-3\right)\left(x-4\right)=0\\ x=3,4\end{array}$

Now, put value $x=4$ in eq. (2):

  $\begin{array}{l}y=7-x\\ y=7-4=3\end{array}$

If the horizontal width is $4$ then the distance from the center to the vertex on the major axis must be half that $a=2$ .

The distance the center to the vertex on the major axis must be half of $3$ at $b=\frac{3}{2}=1.5$ .

Now, if everything is centered at the origin then the equation of the ellipse is written as:

  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

So, put the values:

  $\begin{array}{l}\frac{x^2}{2^2}+\frac{y^2}{{1.5}^2}=1\\ \frac{x^2}{4}+\frac{y^2}{2.25}=1\\ \frac{x^2}{4}+\frac{y^2}{\left(\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)}=1\end{array}$