To determine the standard form of the hyperbola equation given the vertices and asymptotes, let's carefully analyze the problem step by step.
Step 1: Identify the type of hyperbola
The vertices are given as $(0, \pm 4)$. Since the vertices are on the $y$-axis, the hyperbola is vertically oriented. The standard form for a vertically oriented hyperbola is:
\[
\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1
\]
Step 2: Determine the value of \(a\)
The vertices of a hyperbola are located at $(0, \pm a)$ for a vertically oriented hyperbola. Given that the vertices are $(0, \pm 4)$, we have:
\[
a = 4
\]
Thus, \(a^2 = 4^2 = 16\).
Step 3: Use the asymptotes to find \(b\)
The equations of the asymptotes for a vertically oriented hyperbola are given by:
\[
y = \pm \frac{a}{b} x
\]
We are given the asymptotes as \(y = \pm \frac{4}{3} x\). By comparing this with the general form \(y = \pm \frac{a}{b} x\), we can see that:
\[
\frac{a}{b} = \frac{4}{3}
\]
Since \(a = 4\), we substitute \(a\) into the equation:
\[
\frac{4}{b} = \frac{4}{3}
\]
Solving for \(b\):
\[
b = 3
\]
Thus, \(b^2 = 3^2 = 9\).
Step 4: Write the standard form of the hyperbola
Now that we have \(a^2 = 16\) and \(b^2 = 9\), we can write the standard form of the hyperbola:
\[
\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1
\]
Substituting \(a^2 = 16\) and \(b^2 = 9\):
\[
\frac{y^2}{16} - \frac{x^2}{9} = 1
\]
Step 5: Match with the given options
The equation we derived is:
\[
\frac{y^2}{16} - \frac{x^2}{9} = 1
\]
This matches
Option 2.
Final Answer
\[
\boxed{\text{Option 2}}
\]
Parent Tip: Review the logic above to help your child master the concept of hyperbola worksheet.
