1. **State the problem:** Given $a=5+2\sqrt{6}$ and $b=\frac{1}{a}$, find $a^2+b^2$. 2. **Recall the formula:** We want to find $a^2 + b^2$. Since $b=\frac{1}{a}$, we can write $a^2 + b^2 = a^2 + \left(\frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2}$. 3. **Use the identity:** Note that $a^2 + \frac{1}{a^2} = \left(a + \frac{1}{a}\right)^2 - 2$. 4. **Calculate $a + \frac{1}{a}$:** First, find $\frac{1}{a}$: $$\frac{1}{a} = \frac{1}{5 + 2\sqrt{6}}$$ Rationalize the denominator: $$\frac{1}{5 + 2\sqrt{6}} \times \frac{5 - 2\sqrt{6}}{5 - 2\sqrt{6}} = \frac{5 - 2\sqrt{6}}{(5)^2 - (2\sqrt{6})^2} = \frac{5 - 2\sqrt{6}}{25 - 4 \times 6} = \frac{5 - 2\sqrt{6}}{25 - 24} = 5 - 2\sqrt{6}$$ So, $b = \frac{1}{a} = 5 - 2\sqrt{6}$. 5. **Calculate $a + b$:** $$a + b = (5 + 2\sqrt{6}) + (5 - 2\sqrt{6}) = 5 + 5 + 2\sqrt{6} - 2\sqrt{6} = 10$$ 6. **Calculate $a^2 + b^2$ using the identity:** $$a^2 + b^2 = (a + b)^2 - 2ab$$ 7. **Calculate $ab$:** $$ab = (5 + 2\sqrt{6})(5 - 2\sqrt{6}) = 5^2 - (2\sqrt{6})^2 = 25 - 4 \times 6 = 25 - 24 = 1$$ 8. **Substitute values:** $$a^2 + b^2 = 10^2 - 2 \times 1 = 100 - 2 = 98$$ **Final answer:** $$\boxed{98}$$