1. **State the problem:** We have two functions: $$f(x) = \frac{\sqrt{x^2 + k^2}}{x} \quad \text{for } x > 0$$ and $$g(x) = x^2 \quad \text{for } x > 0$$ Given that $$gf(a) = k$$ for $$k > 1$$, find an expression for $$a$$ in terms of $$k$$. 2. **Understand the composition:** The composition $$gf(a)$$ means $$g(f(a))$$. So, $$gf(a) = g\left(f(a)\right) = \left(f(a)\right)^2$$ 3. **Write the expression for $$gf(a)$$:** $$gf(a) = \left(\frac{\sqrt{a^2 + k^2}}{a}\right)^2 = \frac{a^2 + k^2}{a^2}$$ 4. **Set the equation according to the problem:** Given $$gf(a) = k$$, $$\frac{a^2 + k^2}{a^2} = k$$ 5. **Solve for $$a$$:** Multiply both sides by $$a^2$$: $$a^2 + k^2 = k a^2$$ Rewrite: $$a^2 + k^2 = k a^2$$ Bring all terms to one side: $$a^2 - k a^2 + k^2 = 0$$ Factor $$a^2$$: $$a^2(1 - k) + k^2 = 0$$ 6. **Isolate $$a^2$$:** $$a^2(1 - k) = -k^2$$ Divide both sides by $$1 - k$$: $$a^2 = \frac{-k^2}{1 - k}$$ Use \cancel to show simplification: $$a^2 = \frac{-k^2}{\cancel{1 - k}} = \frac{-k^2}{- (k - 1)} = \frac{k^2}{k - 1}$$ 7. **Take the positive square root (since $$a > 0$$):** $$a = \sqrt{\frac{k^2}{k - 1}} = \frac{k}{\sqrt{k - 1}}$$ **Final answer:** $$\boxed{a = \frac{k}{\sqrt{k - 1}}}$$