Chapter 9, Problem 14CR

To determine

To find the Center of the conic $9x^2-y^2-18x+4y-31=0$ .

Answer to Problem 14CR

The hyperbola has a Center as $a.$

  $\left(1,2\right)$ .

Explanation of Solution

Given:

Equation of Conic: $9x^2-y^2-18x+4y-31=0$

Formula Used:

The standard equation for hyperbola- $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ with Center as $\left(h,k\right)$ and foci as $\left(h+c,k\right),\left(h-c,k\right)$ where $c=\sqrt{a^2+b^2}$

Calculation:

Given:

Equation of Conic: $9x^2-y^2-18x+4y-31=0$

The equation can be simplified as:

  $\begin{array}{l}9x^2-y^2-18x+4y-31=0\\ 9x^2-18x-y^2+4y-31=0\\ 9\left(x^2-2x+1\right)-9-\left(y^2-4y+4\right)+4-31=0\\ 9{\left(x-1\right)}^2-{\left(y-2\right)}^2-36=0\\ 9{\left(x-1\right)}^2-{\left(y-2\right)}^2=36\\ \frac{9{\left(x-1\right)}^2-{\left(y-2\right)}^2}{36}=\frac{36}{36}\\ \frac{{\left(x-1\right)}^2}{4}-\frac{{\left(y-2\right)}^2}{36}=1\\ \frac{{\left(x-1\right)}^2}{2^2}-\frac{{\left(y-2\right)}^2}{6^2}=1\end{array}$

Thus, the conic $9x^2-y^2-18x+4y-31=0$ is a hyperbola.

Using the above formula and comparing the above values in the equation of the Hyperbola, we have:

  $h=1$

And $k=2$

Thus,

Center $=$$\left(h,k\right)$

Center $=$$\left(1,2\right)$

Thus, the Center is $\left(1,2\right)$ .

Conclusion:

The hyperbola has a Center as $a.$

  $\left(1,2\right)$ .