Chapter 8.3, Problem 67E

Solution

To solve: To calculate the center of given equation $4x^2-12y^2-16x-72y-44=0$ .

The center is $(2,-3)$ .

Given:

A center calculated by using graphing calculator $4x^2-12y^2-16x-72y-44=0$ .

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 $16$ and for completing the square of $y$ , we added and subtracted the term $108$ ,

  $\begin{array}{c}4x^2-16x-12y^2-72y=44\\ 4\left(x^2-4x+4-4\right)-12\left(y^2+6y+9-9\right)=44\\ $ 4{(x-2)}^2-16-12{(y+3)}^2+108=44$$ 4{(x-2)}^2-12{(y+3)}^2=44-108+16$=-48\end{array}$

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

  $\begin{array}{c}4{(}^x-12{(}^y=-48\\ {\frac{-4(x-2)}{48}}^2-{\frac{12(y+3)}{(-48)}}^2=\frac{-48}{48}\\ {\frac{(y+3)}{4}}^2-\frac{{(x-2)}^2}{12}=1\end{array}$

The center is $(2,-3)$ .

Using the graphing calculator,

  

Thus, the center is $(2,-3)$ .

The correct option is B.