Question

homework assignment 8.2 the hyperbola
score: 1/3 answered: 1/3
question 2
identify a and b for the hyperbola with equation \\( \frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } } = 1 \\).
a =
b =

Explanation:

Step1: Recall hyperbola standard form

The standard form of a horizontal hyperbola is \(\frac{x^2}{a^2}-\frac{y^2}{b^2} = 1\), where \(a\) is the distance from the center to a vertex along the x - axis, and \(b\) is related to the slope of the asymptotes. From the graph, the vertices of the hyperbola are at \((\pm3,0)\), so the distance from the center \((0,0)\) to a vertex is \(a = 3\) (since for the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), the vertices are at \((\pm a,0)\)).

Step2: Determine the slope of asymptotes and find b

The asymptotes of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) have the equation \(y=\pm\frac{b}{a}x\). From the graph, we can see that when \(x = 3\), \(y = 4\) (by looking at the asymptote passing through \((3,4)\) approximately). We know \(a = 3\), and using the asymptote equation \(y=\frac{b}{a}x\), when \(x = 3\) and \(y = 4\), we substitute \(a = 3\) into \(y=\frac{b}{a}x\) to get \(4=\frac{b}{3}\times3\), which simplifies to \(b = 4\). We can also check from the graph's grid. The hyperbola's asymptotes seem to pass through points that give a slope of \(\frac{4}{3}\), and since the slope is \(\frac{b}{a}\) and \(a = 3\), then \(b = 4\).

Answer:

\(a=\boxed{3}\), \(b=\boxed{4}\)