Question

find the equation for the graph of the hyperbola with vertices at (4,0) and (-4,0) and one focus at (-8,0).

Explanation:

Step1: Determine the center of the hyperbola

The center of the hyperbola is the mid - point between the vertices. The vertices are $(4,0)$ and $(-4,0)$. The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. Here, $x_1=4,x_2 = - 4,y_1 = 0,y_2 = 0$, so the center $(h,k)=(0,0)$.

Step2: Find the value of $a$

The distance from the center to a vertex is $a$. The center is $(0,0)$ and a vertex is $(4,0)$, so $a = 4$.

Step3: Find the value of $c$

The distance from the center to a focus is $c$. The center is $(0,0)$ and a focus is $(-8,0)$, so $c = 8$.

Step4: Calculate the value of $b^2$

Use the relationship $c^{2}=a^{2}+b^{2}$ for a hyperbola. Substitute $a = 4$ and $c = 8$ into the formula: $8^{2}=4^{2}+b^{2}$. Then $b^{2}=c^{2}-a^{2}=64 - 16=48$.

Step5: Write the equation of the hyperbola

Since the vertices lie on the $x$ - axis (the $y$ - coordinate of the vertices is 0), the standard form of the hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1$. Substitute $a^{2}=16$ and $b^{2}=48$ into the formula, we get $\frac{x^{2}}{16}-\frac{y^{2}}{48}=1$.

Answer:

$\frac{x^{2}}{16}-\frac{y^{2}}{48}=1$