直线 $\frac{x}{a} + \frac{y}{b} = 1$（其中 $a$ 和 $b$ 为正常数）与 x 轴相交于点 A，与 y 轴相交于点 B。连接 A 和 B 的线段的垂直平分线通过点 $P(-3,-7)$。

1. Find the coordinates of points A and B.

The line $\frac{x}{a} + \frac{y}{b} = 1$ intersects the x-axis at $y=0$, so $x=a$. Thus, $A(a, 0)$.
It intersects the y-axis at $x=0$, so $y=b$. Thus, $B(0, b)$.

2. Apply the property of the perpendicular bisector.

Since $P(-3, -7)$ lies on the perpendicular bisector of $AB$, it is equidistant from $A$ and $B$.
Therefore, $PA = PB$, which implies $PA^2 = PB^2$.

3. Set up the equation using the distance formula.

$(-3 - a)^2 + (-7 - 0)^2 = (-3 - 0)^2 + (-7 - b)^2$
$(a + 3)^2 + 49 = 9 + (b + 7)^2$

4. Expand and simplify to obtain the required equation.

$a^2 + 6a + 9 + 49 = 9 + b^2 + 14b + 49$
$a^2 + 6a = b^2 + 14b$ (Shown)

1. Find the gradient of the line $AB$ and the perpendicular bisector in terms of $a$ and $b$.

The coordinates of $A$ and $B$ are $(a, 0)$ and $(0, b)$ respectively.
Gradient of $AB = \frac{b - 0}{0 - a} = -\frac{b}{a}$
Gradient of the perpendicular bisector $= \frac{-1}{-\frac{b}{a}} = \frac{a}{b}$

2. Form an equation using the given gradient of the perpendicular bisector.

Given the gradient of the perpendicular bisector is 2:
$\frac{a}{b} = 2 \implies a = 2b$

3. Substitute $a = 2b$ into the equation from part (a) to form a quadratic equation in $b$.

From part (a), $a^2 + 6a = b^2 + 14b$.
Substitute $a = 2b$:
$(2b)^2 + 6(2b) = b^2 + 14b$
$4b^2 + 12b = b^2 + 14b$
$3b^2 - 2b = 0$

4. Solve for $b$ and subsequently find $a$, rejecting invalid values.

$b(3b - 2) = 0$
$b = 0$ or $b = \frac{2}{3}$
Since $b$ is a positive constant, we reject $b = 0$.
Therefore, $b = \frac{2}{3}$.
$a = 2\left(\frac{2}{3}\right) = \frac{4}{3}$