Chapter 9.4, Problem 5CYU

a.

To determine

To find: The values of $a$ and $b$ .

Answer to Problem 5CYU

The values of $a$ and $b$ are $240\text{m}$ and $160\text{m}$ .

Explanation of Solution

Given information:

The given figure of an architectural firm is,

  

Concept used:

Write the standard form of the equation of the ellipse whose orientation is along the horizontal axis is,

  $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$

The coordinate of the foci is $\left(h\pm c,k\right)$

The value of $c$ is calculated as,

  $c=\sqrt{a^2-b^2}$

The vertices are at $\left(h\pm a,k\right)$ and the co-vertices are at $\left(h,k\pm b\right)$ .

Write the standard form of the equation of the ellipse whose orientation is along the vertical axis is,

  $\frac{{\left(y-k\right)}^2}{a^2}+\frac{{\left(x-h\right)}^2}{b^2}=1$

The coordinate of the foci is $\left(h,k\pm c\right)$

The vertices are at $\left(h,k\pm a\right)$ and the co-vertices are at $\left(h\pm b,k\right)$ .

The value of $c$ is calculated as,

  $c=\sqrt{a^2-b^2}$

The value of $a$ is calculated as,

  $\begin{array}{c}a=\frac{480}{2}\text{m}\\ =240\text{m}\end{array}$

The value of $b$ is calculated as,

  $\begin{array}{c}b=\frac{320}{2}\text{m}\\ =160\text{m}\end{array}$

Thus, the values of $a$ and $b$ are $240\text{m}$ and $160\text{m}$ .

b.

To determine

To find: The equation of the ellipse.

Answer to Problem 5CYU

The equation for the given ellipse is $\frac{x^2}{57600}+\frac{y^2}{25600}=1$ .

Explanation of Solution

Given information:

The given figure of an architectural firm is,

  

Concept used:

Write the standard form of the equation of the ellipse whose orientation is along the horizontal axis is,

  $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$

The coordinate of the foci is $\left(h\pm c,k\right)$

The value of $c$ is calculated as,

  $c=\sqrt{a^2-b^2}$

The vertices are at $\left(h\pm a,k\right)$ and the co-vertices are at $\left(h,k\pm b\right)$ .

Write the standard form of the equation of the ellipse whose orientation is along the vertical axis is,

  $\frac{{\left(y-k\right)}^2}{a^2}+\frac{{\left(x-h\right)}^2}{b^2}=1$

The coordinate of the foci is $\left(h,k\pm c\right)$

The vertices are at $\left(h,k\pm a\right)$ and the co-vertices are at $\left(h\pm b,k\right)$ .

The value of $c$ is calculated as,

  $c=\sqrt{a^2-b^2}$

The value of $a$ is calculated as,

  $\begin{array}{c}a=\frac{480}{2}\text{m}\\ =240\text{m}\end{array}$

The value of $b$ is calculated as,

  $\begin{array}{c}b=\frac{320}{2}\text{m}\\ =160\text{m}\end{array}$

The standard form of the equation of the given ellipse is,

  $\begin{array}{l}\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1\\ \frac{{\left(x-0\right)}^2}{{\left(240\right)}^2}+\frac{{\left(y-0\right)}^2}{{\left(160\right)}^2}=1\\ \frac{x^2}{57600}+\frac{y^2}{25600}=1\end{array}$

Thus, the equation for the given ellipse is $\frac{x^2}{57600}+\frac{y^2}{25600}=1$ .

c.

To determine

To find: The coordinates of the foci of the ellipse.

Answer to Problem 5CYU

The coordinates of the foci of the ellipse are $\left(\pm 178.88,0\right)$ .

Explanation of Solution

Given information:

The given figure of an architectural firm is,

  

Concept used:

Write the standard form of the equation of the ellipse whose orientation is along the horizontal axis is,

  $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$

The coordinate of the foci is $\left(h\pm c,k\right)$

The value of $c$ is calculated as,

  $c=\sqrt{a^2-b^2}$

The vertices are at $\left(h\pm a,k\right)$ and the co-vertices are at $\left(h,k\pm b\right)$ .

Write the standard form of the equation of the ellipse whose orientation is along the vertical axis is,

  $\frac{{\left(y-k\right)}^2}{a^2}+\frac{{\left(x-h\right)}^2}{b^2}=1$

The coordinate of the foci is $\left(h,k\pm c\right)$

The vertices are at $\left(h,k\pm a\right)$ and the co-vertices are at $\left(h\pm b,k\right)$ .

The value of $c$ is calculated as,

  $c=\sqrt{a^2-b^2}$

The value of $c$ is calculated as,

  $\begin{array}{c}c=\sqrt{a^2-b^2}\\ =\sqrt{{\left(240\right)}^2-{\left(160\right)}^2}\\ =178.88\text{m}\end{array}$

The coordinate of the foci is calculated as,

  $\begin{array}{c}\left(h\pm c,k\right)=\left(0\pm 178.88,0\right)\\ =\left(\pm 178.88,0\right)\end{array}$

Thus, the coordinates of the foci of the ellipse are $\left(\pm 178.88,0\right)$ .