Question

chapter 10 review
score: 40/100 answered: 4/10
● question 5
identify a and b for the hyperbola with equation $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1$.
$a=$
$b=$
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Explanation:

Step1: Recall the standard form of hyperbola

The standard form of a hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1\) has \(a\) as the distance from the center to the vertices along the \(x\) - axis.
From the graph, the vertices of the hyperbola are at \(x=\pm5\). Since for the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), the vertices are \((\pm a,0)\), we have \(a = 5\) (because \(a^{2}=25\) and \(a>0\)).

Step2: Recall the relationship for the asymptotes

The equations of the asymptotes of the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) are \(y=\pm\frac{b}{a}x\).
The slope of the asymptotes can be observed from the graph. The general form of the asymptotes \(y = mx\), and for our hyperbola \(m=\pm\frac{b}{a}\). If we assume the asymptotes pass through the origin and we can use the fact that if we consider a point on the asymptote, say when \(x = 5\), for a hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) with \(a = 5\), the asymptotes \(y=\pm\frac{b}{a}x\). If we assume a "unit - like" relationship (by comparing the ratio of \(y\) and \(x\) for the asymptote). The slope of the asymptote \(m = 1\) (by visual inspection of the graph, since the rise over run for the asymptote is \(1\) when \(a = 5\)). Since \(m=\frac{b}{a}\) and \(a = 5\) and \(m = 1\), we get \(b=a\).

Answer:

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