Chapter 8.2, Problem 69E

Solution

To calculate the center of given equation $9x^2+4y^2-72x-24y+144=0$ .

The center is $(4,3)$ .

Given:

A center calculated by using graphing calculator $9x^2+4y^2-72x-24y+144=0$

Concept Used:

The standard equation of the center can be written as:

Calculation:

As per the given problem,

For making the completing the square term, we have added and subtracted the term $144$ and for completing the square of $y$ , we added and subtracted the term $36$ ,

  $\begin{array}{c}9x^2-72x+144-144+4y^2-24y+36-36+144=0\\ \left(9x^2-72x+144\right)+\left(4y^2-24y+36\right)+\left(-144-36+144\right)=0\\ \left(9x^2-72x+144\right)+\left(4y^2-24y+36\right)+\left(-36\right)=0\\ $ {(3x-12)}^2+{(2y-6)}^2=36$\end{array}$

For the center, we can equate the values to zero as,

  $\begin{array}{c}3x-12=0\\ x=4\\ 2y-6=0\\ y=\frac{6}{2}\\ y=3\end{array}$

The center is $(4,3)$

Using the graphing calculator,

  

Thus, the center is $(4,3)$ .

The correct option is B.