Chapter 10.4, Problem 51PPSE

Solution

To find: The equation in standard form of an ellipse with center at the origin.

The required equation of the ellipse is $\frac{x^2}{\frac{2809}{4}}+\frac{y^2}{\frac{841}{4}}=1$ .

Given information: The height of the ellipse is $29$ and width is $53$ .

Formula used: The equation of the ellipse is,

  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ; $a>b>0$ .

Calculation:

The height of the ellipse is $29$ and width is $53$ .

This implies that $a=\frac{53}{2}$ and $b=\frac{29}{2}$ .

Put, the values of $a$ and $b$ in $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ,

  $\begin{array}{c}\frac{x^2}{{\left(\frac{53}{2}\right)}^2}+\frac{y^2}{{\left(\frac{29}{2}\right)}^2}=1\\ \frac{x^2}{\frac{2809}{4}}+\frac{y^2}{\frac{841}{4}}=1\end{array}$

Hence, the required equation of the ellipse is $\frac{x^2}{\frac{2809}{4}}+\frac{y^2}{\frac{841}{4}}=1$ .