Chapter 9, Problem 12CR

To determine

To find the focus of the hyperbola $8x^2-3y^2=48$ .

Answer to Problem 12CR

The Focus is $b.$

  $\left(\sqrt{22},0\right)$

Explanation of Solution

Given:

Equation of Hyperbola: $8x^2-3y^2=48$

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 Hyperbola: $8x^2-3y^2=48$

The equation can be simplified as:

  $\begin{array}{l}8x^2-3y^2=48\\ \frac{8x^2-3y^2}{48}=\frac{48}{48}\\ \frac{8x^2}{48}-\frac{3y^2}{48}=1\\ \frac{x^2}{6}-\frac{y^2}{16}=1\\ \frac{x^2}{{\left(\sqrt{6}\right)}^2}-\frac{y^2}{4^2}=1\end{array}$

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

  $h=0$

And $k=0$

Thus,

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

Center $=$$\left(0,0\right)$

  $a=\sqrt{6}$ and $b=4$

  $c=\sqrt{a^2+b^2}$

  $\begin{array}{l}c=\sqrt{6+16}\\ c=\sqrt{22}\end{array}$

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

Foci $=$$\left(0+\sqrt{22},0\right),\left(0-\sqrt{22},0\right)$

Foci $=$$\left(\sqrt{22},0\right),\left(-\sqrt{22},0\right)$

Thus, the Focus is $b.$

  $\left(\sqrt{22},0\right)$

Conclusion:

The Focus is $b.$

  $\left(\sqrt{22},0\right)$