Given that,
- $(a+3)^{2} : b^{2} = 9 : 1 \quad \longrightarrow (1)$
- $(a-1)^{2} : (b-1)^{2} = 4 : 1 \quad \longrightarrow (2)$
From equation $(1), \frac{(a+3)^{2}}{b^{2}} = \frac{9}{1}$
$\Rightarrow \left(\frac{a+3}{b}\right)^{2} = \left(\frac{3}{1}\right)^{2}$
$\Rightarrow \frac{a+3}{b} = \pm \frac{3}{1}$
$\Rightarrow a+3 = 3b \quad \text{ or } \quad a+3 = -3b \longrightarrow (3) $
From equation $(2),\frac{(a-1)^{2}}{(b-1)^{2}} = \frac{4}{1}$
$\Rightarrow \left(\frac{a-1}{b-1}\right)^{2} = \left(\frac{2}{1}\right)^{2}$
$\Rightarrow \frac{a-1}{b-1} = \pm 2$
$\Rightarrow a - 1 = 2b-2 \quad \text{or} \quad a-1 = 2 - 2b \longrightarrow (4) $
Analyze the 4 cases
Case 1: $a + 3 = 3b$ and $a - 1 = 2(b - 1)$
From $a + 3 = 3b$:
$$a = 3b - 3$$
Substituting into $a - 1 = 2(b - 1)$:
$$3b - 3 - 1 = 2b - 2$$ $$3b - 4 = 2b - 2$$ $$b = 2$$
Substituting $b = 2$ into $a = 3b - 3$:
$$a = 3(2) - 3 = 6 - 3 = 3$$
With $a = 3$ and $b = 2$, we calculate:
$$\frac{a^2}{b^2} = \frac{3^2}{2^2} = \frac{9}{4}$$
Case 2: $a + 3 = -3b$ and $a - 1 = 2(b - 1)$
From $a + 3 = -3b$:
$$a = -3b - 3$$
Substituting into $a - 1 = 2(b - 1)$:
$$-3b - 3 - 1 = 2b - 2$$ $$-3b - 4 = 2b - 2$$ $$-5b = 2$$
$b = -\frac{2}{5}$, which is not an integer, so this case is invalid.
Case 3: $a + 3 = 3b$ and $a - 1 = -2(b - 1)$
From $a + 3 = 3b$:
$$a = 3b - 3$$
Substituting into $a - 1 = -2(b - 1)$:
$$3b - 3 - 1 = -2(b - 1)$$ $$3b - 4 = -2b + 2$$ $$5b = 6$$
$b = \frac{6}{5}$, which is not an integer, so this case is invalid.
Case 4: $a + 3 = -3b$ and $a - 1 = -2(b - 1)$
From $a + 3 = -3b$:
$$a = -3b - 3$$
Substituting into $a - 1 = -2(b - 1)$:
$$-3b - 3 - 1 = -2b + 2$$ $$-3b - 4 = -2b + 2$$ $$-b = 6 \quad \Rightarrow \quad b = -6$$
Substituting $b = -6$ into $a = -3b - 3$:
$$a = -3(-6) - 3 = 18 - 3 = 15$$
With $a = 15$ and $b = -6$, we calculate:
$$\frac{a^2}{b^2} = \frac{15^2}{6^2} = \frac{225}{36} = \frac{25}{4}$$
Conclusion
The correct ratio $\frac{a^2}{b^2}$ is:
$$\boxed{25:4}$$
