Chapter 9, Problem 10CT

To determine

To identify the conic $x^2-4y^2-4x-8y-4=0$ and find it’s Center and foci.

Answer to Problem 10CT

The hyperbola has a Center as $\left(2,-1\right)$ and Foci as $\left(2+\sqrt{5},-1\right),\left(2-\sqrt{5},-1\right)$ .

Explanation of Solution

Given:

Equation of Conic: $x^2-4y^2-4x-8y-4=0$

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: $x^2-4y^2-4x-8y-4=0$

The equation can be simplified as:

  $\begin{array}{l}{x}^2-4y^2-4x-8y-4=0\\ x^2-4x+4-4-4y^2-8y-4+4-4=0\\ \left(x^2-4x+4\right)-4-\left(4y^2+8y+4\right)+4-4=0\\ {\left(x-2\right)}^2-{\left(2y+2\right)}^2-4=0\\ {\left(x-2\right)}^2-4{\left(y+1\right)}^2=4\\ \frac{{\left(x-2\right)}^2-4{\left(y+1\right)}^2}{4}=\frac{4}{4}\\ \frac{{\left(x-2\right)}^2}{4}-\frac{{\left(y+1\right)}^2}{1}=1\end{array}$

Thus, the conic $x^2-4y^2-4x-8y-4=0$ is a hyperbola.

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

  $h=2$

And $k=-1$

Thus,

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

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

  $a=2$ and $b=1$

  $c=\sqrt{a^2+b^2}$

  $\begin{array}{l}c=\sqrt{4+1}\\ c=\sqrt{5}\end{array}$

Now, Foci $=$$\left(h+c,k\right),\left(h-c,k\right)$

Foci $=$$\left(2+\sqrt{5},-1\right),\left(2-\sqrt{5},-1\right)$

Thus, the Center is $\left(2,-1\right)$ and Foci is $\left(2+\sqrt{5},-1\right),\left(2-\sqrt{5},-1\right)$ .

Conclusion:

The hyperbola has a Center as $\left(2,-1\right)$ and Foci as $\left(2+\sqrt{5},-1\right),\left(2-\sqrt{5},-1\right)$ .