Okay, so I need to find the equation of a hyperbola given the foci F1 = (10, 2) and F2 = (-16, 2), and the condition that the absolute difference of the distances from any point P on the hyperbola to these foci is 24. Then, once I have the equation in the standard form, I need to find the sum h + k + a + b. Let me start by recalling some basic properties of hyperbolas.

First, the standard form of a hyperbola is either \(\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\) or \(\frac{(y - k)^2}{b^2} - \frac{(x - h)^2}{a^2} = 1\), depending on whether it's opening horizontally or vertically. Since the foci are given as (10, 2) and (-16, 2), both have the same y-coordinate, which is 2. That suggests that the hyperbola opens horizontally, because the foci are aligned horizontally. Therefore, the standard form we need to use is the first one: \(\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\). So that's good, we can focus on that form.

Now, for hyperbolas, the distance between the two foci is 2c, so first I should find the center of the hyperbola, which is the midpoint between the two foci. The midpoint formula is ((x1 + x2)/2, (y1 + y2)/2). Let me compute that.

F1 is (10, 2), F2 is (-16, 2). The x-coordinate of the midpoint is (10 + (-16))/2 = (-6)/2 = -3. The y-coordinate is (2 + 2)/2 = 2. So the center (h, k) is (-3, 2). So h = -3, k = 2. That's part of the equation.

Next, the distance between the foci is 2c. The distance between F1 and F2 can be calculated using the distance formula. Let's do that. The distance between (10, 2) and (-16, 2) is sqrt[(10 - (-16))² + (2 - 2)²] = sqrt[(26)² + 0] = 26. Therefore, 2c = 26, so c = 13.

The problem states that |PF1 - PF2| = 24. In hyperbola terminology, this value 24 is equal to 2a, where a is the distance from the center to the vertices along the transverse axis. So 2a = 24, which means a = 12. Therefore, a = 12.

Now, in hyperbolas, the relationship between a, b, and c is given by c² = a² + b². We know c = 13 and a = 12, so we can solve for b.

Calculating c²: 13² = 169.

Calculating a²: 12² = 144.

Therefore, b² = c² - a² = 169 - 144 = 25. So b² = 25, which means b = 5. Since b is a length, we take the positive square root, so b = 5.

Now, putting it all together into the standard form equation. The center is (-3, 2), so h = -3, k = 2. a² = 144, b² = 25. Therefore, the equation is:

\[
\frac{(x - (-3))^2}{144} - \frac{(y - 2)^2}{25} = 1
\]

Simplifying the x-term: (x + 3)². So the equation becomes:

\[
\frac{(x + 3)^2}{144} - \frac{(y - 2)^2}{25} = 1
\]

So now, we need to find h + k + a + b. Let's recall h = -3, k = 2, a = 12, b = 5. Adding them together: (-3) + 2 + 12 + 5.

Calculating step by step:

-3 + 2 = -1

-1 + 12 = 11

11 + 5 = 16

Therefore, the sum h + k + a + b is 16.

Wait, let me verify each step again to make sure I didn't make a mistake.

First, the center: midpoint between (10, 2) and (-16, 2). The x-coordinate: (10 + (-16))/2 = (-6)/2 = -3. Correct. Y-coordinate: (2 + 2)/2 = 2. Correct. So center is (-3, 2). So h = -3, k = 2. Correct.

Distance between foci: sqrt[(10 - (-16))² + (2 - 2)²] = sqrt[26²] = 26. So 2c = 26, c = 13. Correct.

Given that |PF1 - PF2| = 24 = 2a, so a = 12. Correct.

Then using c² = a² + b²: 169 = 144 + b², so b² = 25, so b = 5. Correct.

Then the equation: \(\frac{(x + 3)^2}{144} - \frac{(y - 2)^2}{25} = 1\). Correct.

Sum h + k + a + b: (-3) + 2 + 12 + 5. Let's compute again:

-3 + 2 = -1

-1 + 12 = 11

11 + 5 = 16. Correct.

So yes, the final answer is 16.

But just to be thorough, let me check if the hyperbola equation is correctly oriented. Since the foci are aligned horizontally (same y-coordinate), the transverse axis is horizontal, so the hyperbola opens left and right. The standard form is indeed \(\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\), which matches what we have. Therefore, all steps are correct.

Therefore, the answer is 16.
So, the final answer is $\boxed{16}$.
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