Answer:
[tex]\frac{x^2}{12}-\frac{y^2}{4}=1[/tex]
Step-by-step explanation:
Since our foci are located on the x-axis, then our major axis is going to be the horizontal transverse axis of the hyperbola:
Formula for hyperbola with horizontal transverse axis centered at origin
- [tex]\frac{x^2}{a^2}-\frac{y^2}{b^2}=1[/tex]
- Directrices -> [tex]x=\pm\frac{a^2}{c}[/tex]
- Foci -> [tex](\pm c,0)[/tex] where [tex]a^2+b^2=c^2[/tex]
- [tex]a>b[/tex]
Since we are given our directrices of [tex]x=\pm3[/tex] and foci of [tex](\pm4,0)[/tex], then we can set up the directrices equation to solve for [tex]a^2[/tex]:
[tex]x=\pm\frac{a^2}{c}\\ \\\pm3=\pm\frac{a^2}{4}\\ \\12=a^2[/tex]
Now we can determine [tex]b^2[/tex] to complete our equation for the hyperbola:
[tex]a^2+b^2=c^2\\\\12+b^2=4^2\\\\12+b^2=16\\\\b^2=4[/tex]
Therefore, our equation for our hyperbola is [tex]\frac{x^2}{12}-\frac{y^2}{4}=1[/tex]
