Chapter 8, Problem 8TP

Solution

To calculate: Without using the areas formula, the ellipses' representations of their areas.

They both share a space.

Given information: This leads to the conclusion that these ellipses are identical. Just the first one translated so that the center is in the second one.

Calculation:

The ellipse

  $4x^2+6y^2=48$

Can be expressed as

  $\begin{array}{r}\hfill \frac{4x^2}{48}+\frac{6y^2}{48}=1\\ \hfill \frac{x^2}{12}+\frac{y^2}{8}=1\end{array}$

A statement that its half-axes:

  $\begin{array}{c}a=\sqrt{12}\\ =2\sqrt{3}\end{array}$

And

  $\begin{array}{c}b=\sqrt{8}\\ =2\sqrt{2}\end{array}$ .

The half-axes of the ellipse

  $\begin{array}{c}a=\sqrt{12}\\ =2\sqrt{3}\end{array}$

  $$

  $\frac{{(x-5)}^2}{12}+\frac{{(y+7)}^2}{8}=1$

Are equivalent. This leads us to the conclusion that these ellipses are identical. The second one has the same area because it is simply the first one translated so that its center is $(5,-7)$ .