Question

identify a and b for the hyperbola with equation $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1$.

Explanation:

Step1: Recall the standard form of hyperbola

The standard form of a hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1\) (\(a>0,b>0\)). Here, \(a\) is the distance from the center to the vertices along the \(x\) - axis, and \(b\) is related to the slope of the asymptotes \(y=\pm\frac{b}{a}x\).

Step2: Determine the value of \(a\)

From the graph, the distance from the center \((0,0)\) to the vertices (on the \(x\) - axis) is \(a\). Since the vertices are at \(x = \pm5\), then \(a^{2}=25\), so \(a = 5\) (because \(a>0\)).

Step3: Determine the value of \(b\)

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, the asymptotes have a slope of \(\pm1\) (since the lines pass through the origin and for example, if we consider a point \((a,b)=(5,b)\) on the asymptote \(y = \frac{b}{a}x\), and the slope of the asymptote is \(1\) (by visual inspection of the symmetry and the ratio of \(x\) and \(y\) - intercepts of the asymptotes). Since \(a = 5\) and \(y=\pm\frac{b}{a}x\) with slope \(1\), then \(\frac{b}{a}=1\). Substituting \(a = 5\), we get \(b=5\) (because \(b>0\)).

Answer:

\(a = 5\), \(b = 5\)