Chapter 9, Problem 40RE

To determine

Calculate the longest distance across the pool, the shortest distance, and the distance between the foci.

Answer to Problem 40RE

The longest distance , shortest distance and the distance between the foci are $36,28and16\sqrt{2}.$

Explanation of Solution

Given: It is given in the question that the equation of the elliptical shape pool is $\frac{x^2}{324}+\frac{y^2}{196}=1$

Concept Used:

In this, use the concept of ellipse and also use the formula of $c=\sqrt{a^2-b^2}$ .

Calculation: Here, the equation is $\frac{x^2}{324}+\frac{y^2}{196}=1$ .

Now, it has to calculate a and b, and it is also know that $2a$ is the longest, and $2b$ is the shortest distance.

  $\begin{array}{l}{a}^2=324\\ a=\sqrt{324}=18\end{array}$

  $\begin{array}{l}{b}^2=196\\ b=\sqrt{196}=14\end{array}$

Now,

  $2a=2\times 18=36$ , $2b=2\times 14=28$

Now, use the formula to find c,

  $\begin{array}{l}c=\sqrt{a^2-b^2}\\ c=\sqrt{{18}^2-{14}^2}\\ c=\sqrt{324-196}\\ c=\sqrt{128}\\ c=8\sqrt{2}\end{array}$

So, $2c$ is the distance between the foci.

  $2c=2\times 8\sqrt{2}=16\sqrt{2}$

Conclusion:

The longest distance , shortest distance and the distance between the foci are $36,28and16\sqrt{2}.$