Step 1: Identify the general term in the binomial expansion. The general term $T_{k+1}$ in the expansion of $(a+b)^n$ is given by the formula: $$T_{k+1} = \binom{n}{k} a^{n-k} b^k$$ In this problem, we have $(Y^2X^2 + X^4)^{45}$. So, $a = Y^2X^2$, $b = X^4$, and $n = 45$. Step 2: Substitute the values into the general term formula. $$T_{k+1} = \binom{45}{k} (Y^2X^2)^{45-k} (X^4)^k$$ Step 3: Simplify the expression to find the exponent of $X$. $$T_{k+1} = \binom{45}{k} (Y^2)^{45-k} (X^2)^{45-k} (X^4)^k$$ $$T_{k+1} = \binom{45}{k} Y^{2(45-k)} X^{2(45-k)} X^{4k}$$ $$T_{k+1} = \binom{45}{k} Y^{90-2k} X^{90-2k+4k}$$ $$T_{k+1} = \binom{45}{k} Y^{90-2k} X^{90+2k}$$ Step 4: Set the exponent of $X$ equal to $100$ and solve for $k$. We are looking for terms in $X^{100}$, so we set the exponent of $X$ to $100$: $$90+2k = 100$$ $$2k = 100 - 90$$ $$2k = 10$$ $$k = \frac{10}{2}$$ $$k = 5$$ Since $k=5$ is an integer and $0 \le 5 \le 45$, this is a valid value for $k$. Step 5: Substitute $k=5$ back into the general term expression. $$T_{5+1} = T_6 = \binom{45}{5} Y^{90-2(5)} X^{90+2(5)}$$ $$T_6 = \binom{45}{5} Y^{90-10} X^{90+10}$$ $$T_6 = \binom{45}{5} Y^{80} X^{100}$$ Step 6: Calculate the binomial coefficient $\binom{45}{5}$. $$\binom{45}{5} = \frac{45!}{5!(45-5)!} = \frac{45!}{5!40!} = \frac{45 \times 44 \times 43 \times 42 \times 41}{5 \times 4 \times 3 \times 2 \times 1}$$ $$\binom{45}{5} = \frac{45}{5 \times 3} \times \frac{44}{4 \times 2} \times 43 \times 42 \times 41$$ $$\binom{45}{5} = (3) \times (11) \times 43 \times (7) \times 41$$ $$\binom{45}{5} = 9 \times 11 \times 43 \times 7 \times 41$$ $$\binom{45}{5} = 99 \times 301 \times 41$$ $$\binom{45}{5} = 29799 \times 41$$ $$\binom{45}{5} = 1,221,759$$ The term containing $X^{100}$ is: $$\boxed{1,221,759 Y^{80} X^{100}}$$