Question

a hyperbola with equation of \\( \frac{(x + 5)^{2}}{a^{2}}-\frac{(y + 7)^{2}}{b^{2}} = 1 \\) has an asymptote with equation of \\( y=\frac{9}{4}x+\frac{17}{4} \\). find the smallest possible whole number values for \\( a \\) and \\( b \\).

Explanation:

Step1: Recall the asymptote formula for hyperbola

For a hyperbola \(\frac{(x - h)^{2}}{a^{2}}-\frac{(y - k)^{2}}{b^{2}} = 1\), the asymptotes are \(y-k=\pm\frac{b}{a}(x - h)\). In the given hyperbola \(\frac{(x + 5)^{2}}{a^{2}}-\frac{(y+7)^{2}}{b^{2}}=1\), \(h=-5\), \(k = - 7\). The given asymptote is \(y=\frac{9}{4}x+\frac{17}{4}\), and its slope - intercept form \(y+7=\frac{9}{4}(x + 5)\) (since \(y+7=\frac{b}{a}(x + 5)\) for one of the asymptotes).

Step2: Compare the slopes

Comparing \(y+7=\frac{b}{a}(x + 5)\) with \(y+7=\frac{9}{4}(x + 5)\), we get \(\frac{b}{a}=\frac{9}{4}\).

Since \(a\) and \(b\) are whole - numbers, the smallest non - negative whole - number values occur when \(a = 4\) and \(b=9\) (by cross - multiplying \(\frac{b}{a}=\frac{9}{4}\) to get \(4b=9a\), and looking for the smallest positive integers \(a\) and \(b\) that satisfy the equation).

Answer:

\(a = 4\), \(b = 9\)