Question

Which is the equation of a hyperbola with directrices at x = ±3 and foci at (4, 0) and (−4, 0)?​

Answer 1

Answer:

$\frac{x^2}{12}-\frac{y^2}{4}=1$

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

• $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
• Directrices -> $x=\pm\frac{a^2}{c}$
• Foci -> $(\pm c,0)$ where $a^2+b^2=c^2$
• $a>b$

Since we are given our directrices of $x=\pm3$ and foci of $(\pm4,0)$, then we can set up the directrices equation to solve for $a^2$:

$x=\pm\frac{a^2}{c}\\ \\\pm3=\pm\frac{a^2}{4}\\ \\12=a^2$

Now we can determine $b^2$ to complete our equation for the hyperbola:

$a^2+b^2=c^2\\\\12+b^2=4^2\\\\12+b^2=16\\\\b^2=4$

Therefore, our equation for our hyperbola is $\frac{x^2}{12}-\frac{y^2}{4}=1$