Chapter 9, Problem 37RE

To determine

Calculate the standard form of the equation of the ellipse.

Answer to Problem 37RE

The standard from of the equation is $\frac{{(x-2)}^2}{25}+\frac{y^2}{21}=1$ .

Explanation of Solution

Given:

It is given in the question that the $vertices:(-3,0),(7,0);Foci:(0,0),(4,0)$ .

Concept Used:

In this, use the concept of the ellipse and use the equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ .

Calculation: Since, the given vertices are $(-3,0)$ and $(7,0)$ , half of the length of the horizontal axis is $5$ . Since the vertices are on the same line as the foci, this length is half of the length of the major axis. The center of the ellipse is at $(2,0)$ . Half the distance between the foci is $2$ .Now, it has to be calculate the length of the horizontal axis.The square of the distance between one of the foci and the center plus the square of half of the minor axis length is equal to the square of half of the major axis length.

  $\begin{array}{l}b=\sqrt{5^2-2^2}\\ b=\sqrt{25-4}\\ b=\sqrt{21}\end{array}$

And the $a=5$ ,

Now, use the equation,

  $\begin{array}{l}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(ithascentretoo)\\ \frac{{(x-2)}^2}{5^2}+\frac{{(y-0)}^2}{{\sqrt{21}}^2}=1\\ \frac{{(x-2)}^2}{25}+\frac{y^2}{21}=1\end{array}$

Conclusion:

The standard equation is $\frac{{(x-2)}^2}{25}+\frac{y^2}{21}=1$ .