The standard form of the ellipse with the foci on the x-axis
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\ a+b=20$$
Given the formula for the area of ellipse is
$$A=\pi ab$$
a) How do I find the area of the ellipse as a function of $a$?
b) I want to know the equation of an ellipse given the area of $264\ cm^2$ c) I was asked to complete the table based from part(a), and make a guess about the shape of the ellipse
| a | 8 | 9 | 10 | 11 |
| A | _ | _ | _ | _ |
Where a is the function, and A is the area.
On
First question
Since $$a+b=20$$ you have that $$b=20-a$$. Therefore the area A is given by
$$ A = \pi a\left( {20 - a} \right) $$
Second question
You have to solve the second degree equation $$ \pi a\left( {20 - a} \right) = 264 $$ which turn to be $$ a^2 - 20a + \frac{{264}} {\pi } = 0 $$ You get two solutions
$$ \begin{gathered} a_1 = 10 + \sqrt {100 - \frac{{264}} {\pi }} \hfill \\ \hfill \\ a_2 = 10 - \sqrt {100 - \frac{{264}} {\pi }} \hfill \\ \end{gathered} $$ Geometrically this means that you get two ellipses: one of them has the major axis on x-axis the other on the y-axis.
Third question You have only to substitute and find the values of A. For the shape
For part a), you can use the fact that $b=20-a$ (or that $a = 20-b$).
For the table in part b), compare the ellipse to $x^2+y^2=1$. Now, using the transformation $x \rightarrow \frac{x}{8}$, the ellipse is squished by a factor of $\frac{1}{8}$ in the $x$-axis, or stretched by a factor of $8$.
In addition, by the transformation $y \rightarrow \frac{y}{12}$, the ellipse is stretched by a factor of $12$ in the $y$-axis. Therefore, the ellipse will be taller than it is wide. You should fill in the table from here.
To find the ellipse that has an area of $264 \text{cm}^2$, you can use $b=20-a$ to get $\pi a(20-a) = 264$. This is a simple quadratic equation. Now solve for $a$, and then using $b=20-a$ you can solve for $b$.