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Let's solve the problem step by step.

### Problem:
We need to find the equation of the tangent line to the curve \( f(x) = \sqrt{ax} + \sqrt{ax} \) at the point where \( x = 1 \). We are asked to give the y-intercept of this tangent line. Additionally, the value of \( a \) is given as 18.

### Step 1: Simplify the function
The function given is:
\[
f(x) = \sqrt{ax} + \sqrt{ax}
\]
This can be simplified as:
\[
f(x) = 2\sqrt{ax}
\]

### Step 2: Find the derivative \( f'(x) \)
To find the equation of the tangent line, we first need to calculate the derivative of \( f(x) \):
\[
f'(x) = \frac{d}{dx} \left( 2\sqrt{ax} \right)
\]
Using the chain rule, we have:
\[
f'(x) = 2 \cdot \frac{1}{2} \cdot (ax)^{-\frac{1}{2}} \cdot a
\]
Simplifying, we get:
\[
f'(x) = \frac{a}{\sqrt{ax}}
\]
Substituting \( a = 18 \), the derivative becomes:
\[
f'(x) = \frac{18}{\sqrt{18x}}
\]
Further simplification gives:
\[
f'(x) = \frac{18}{\sqrt{18} \cdot \sqrt{x}} = \frac{18}{3\sqrt{2x}} = \frac{6}{\sqrt{2x}}
\]

### Step 3: Evaluate the derivative at \( x = 1 \)
Now, we find the slope of the tangent line at \( x = 1 \):
\[
f'(1) = \frac{6}{\sqrt{2 \cdot 1}} = \frac{6}{\sqrt{2}} = \frac{6\sqrt{2}}{2} = 3\sqrt{2}
\]
So, the slope of the tangent line is \( m = 3\sqrt{2} \).

### Step 4: Find the y-coordinate at \( x = 1 \)
We now calculate \( f(1) \) to find the y-coordinate of the point where the tangent touches the curve:
\[
f(1) = 2\sqrt{18 \cdot 1} = 2\sqrt{18} = 2 \cdot 3\sqrt{2} = 6\sqrt{2}
\]
So, the point of tangency is \( (1, 6\sqrt{2}) \).

### Step 5: Equation of the tangent line
The equation of the tangent line is given by:
\[
y - y_1 = m(x - x_1)
\]
Substituting \( m = 3\sqrt{2} \), \( x_1 = 1 \), and \( y_1 = 6\sqrt{2} \), we get:
\[
y - 6\sqrt{2} = 3\sqrt{2}(x - 1)
\]
Expanding this, we obtain:
\[
y = 3\sqrt{2}x - 3\sqrt{2} + 6\sqrt{2}
\]
\[
y = 3\sqrt{2}x + 3\sqrt{2}
\]

### Step 6: Find the y-intercept
The y-intercept occurs when \( x = 0 \):
\[
y = 3\sqrt{2}(0) + 3\sqrt{2} = 3\sqrt{2}
\]

### Final Answer:
The y-intercept of the tangent line is \( \boxed{3\sqrt{2}} \).

Would you like further details or have any questions?

Here are some related questions for further exploration:
1. How do you find the tangent line to a curve at a given point?
2. What role does the derivative play in determining the slope of the tangent line?
3. How can you calculate the y-intercept of a line given its equation?
4. What are the applications of tangent lines in real-world problems?
5. How does the chain rule simplify the process of differentiation?

**Tip:** Understanding the geometric interpretation of the derivative as the slope of the tangent line can greatly enhance your comprehension of calculus.

Learn how to find the equation of the tangent line to the curve f(x) = 2√(18x) at the point x = 1. This problem involves calculus, derivatives using the chain rule, and finding the y-intercept of the tangent line. Suitable for advanced high school or college students.

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