If a and b are integers of opposite signs such that $\frac{{(a+3)}^2}{b^2}=\frac{9}{1}$ and $\frac{{(a-1)}^2}{{(b-1)}^2}=\frac{4}{1}$, then the ratio a : b is

The correct option is D.

We are given two equations involving two integers $a$ and $b$ of opposite signs.

First equation:
$\frac{{(a+3)}^2}{b^2}=9$
Taking the square root on both sides, we get:
$a+3=\pm 3b$

Second equation:
$\frac{{(a-1)}^2}{{(b-1)}^2}=4$
Taking the square root on both sides, we get:
$a-1=\pm 2(b-1)$

Let's analyze the cases based on the opposite signs of $a$ and $b$.

Case 1: $a+3=3b$
If $a+3=3b$, then $a=3b-3$.
Since $a$ and $b$ must have opposite signs, let's test the second equation with $a-1=2(b-1)$:
$3b-3-1=2b-2$
$3b-4=2b-2$
$b=2$
Then $a=3\left(2\right)-3=3$. Here, both $a$ and $b$ are positive, which violates the condition that they are of opposite signs.

Now let's test the second equation with $a-1=-2(b-1)$:
$3b-3-1=-2b+2$
$3b-4=-2b+2$
$5b=6$
This does not yield an integer value for $b$.

Case 2: $a+3=-3b$
Then $a=-3b-3$.
Let's test the second equation with $a-1=2(b-1)$:
$-3b-3-1=2b-2$
$-3b-4=2b-2$
$-5b=2$
This does not yield an integer value for $b$.

Now let's test the second equation with $a-1=-2(b-1)$:
$-3b-3-1=-2b+2$
$-3b-4=-2b+2$
$-b=6$
$b=-6$
Substituting $b=-6$ into $a=-3b-3$:
$a=-3(-6)-3=18-3=15$

Here, $a=15$ (positive) and $b=-6$ (negative) are integers of opposite signs, which satisfies all the conditions.

The ratio of their squares is:
$a^2:b^2={15}^2:{(-6)}^2=225:36=25:4$