Find Sum of Squares of Interval Boundaries for Function Parameter

To find the set of all possible values of the parameter $a$, we analyze the given equation of the ellipse:

$\frac{x^2}{f\left(a^2 + 7a + 3\right)} + \frac{y^2}{f(3a + 15)} = 1$

For this equation to represent an ellipse with its major axis along the $y$-axis, the coefficient of $y^2$ in the standard form must correspond to a larger denominator than that of $x^2$. That is, the semi-major axis squared along the $y$-axis must be strictly greater than the semi-minor axis squared along the $x$-axis:

$f(3a + 15) > f\left(a^2 + 7a + 3\right)$

We are given that $f(x)$ is a strictly decreasing positive function on $\mathbf{R}$. By the definition of a strictly decreasing function, for any $x_1, x_2 \in \mathbf{R}$:

$f(x_1) > f(x_2) \iff x_1 < x_2$

Applying this property to our inequality, we get:

$3a + 15 < a^2 + 7a + 3$

Rearranging the terms to one side of the inequality yields:

$a^2 + 4a - 12 > 0$

Factoring the quadratic expression:

$(a + 6)(a - 2) > 0$

Solving this inequality gives:

$a \in (-\infty, -6) \cup (2, \infty)$

In set-difference notation, this solution set can be expressed as:

$\mathbf{R} - [-6, 2]$

Comparing this set with the given form $\mathbf{R} - [\alpha, \beta]$, we identify:

$\alpha = -6 \quad \text{and} \quad \beta = 2$

Finally, we calculate $\alpha^2 + \beta^2$:

$\alpha^2 + \beta^2 = (-6)^2 + (2)^2 = 36 + 4 = 40$

Thus, the correct option is B.