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The problem given is:

1. \( y \propto \frac{1}{x^2} \)
2. \( x = 4, y = 2 \)

### Step-by-Step Solution

1. **Interpret the Proportional Relationship**:
   Since \( y \propto \frac{1}{x^2} \), we can write:
   \[
   y = \frac{k}{x^2}
   \]
   where \( k \) is the constant of proportionality.

2. **Substitute the Values of \( x \) and \( y \) to Find \( k \)**:
   Given \( x = 4 \) and \( y = 2 \), we substitute these values into the equation:
   \[
   2 = \frac{k}{4^2}
   \]
   Simplifying, we get:
   \[
   2 = \frac{k}{16}
   \]
   Multiplying both sides by 16 to solve for \( k \):
   \[
   k = 2 \times 16 = 32
   \]

3. **Rewrite the Equation with the Found Constant \( k \)**:
   Now that we know \( k = 32 \), we can rewrite the equation as:
   \[
   y = \frac{32}{x^2}
   \]

### Final Answer
The equation that describes the relationship between \( y \) and \( x \) is:
\[
y = \frac{32}{x^2}
\]

Would you like more details or have any questions?

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Here are some related questions for further understanding:

1. What would be the value of \( y \) if \( x = 5 \) using the found equation?
2. How does the graph of \( y = \frac{32}{x^2} \) look, and what is its general shape?
3. If \( y \propto \frac{1}{x^n} \), how does changing \( n \) affect the behavior of \( y \) as \( x \) increases?
4. Can we solve for \( x \) in terms of \( y \) from the equation \( y = \frac{32}{x^2} \)?
5. What would happen if \( y \) were directly proportional to \( x^2 \) instead of inversely proportional?

**Tip:** In inverse proportionality, as \( x \) increases, \( y \) decreases, following a specific rate based on the power in the equation.

Learn how to determine the constant of proportionality in an inverse square relationship, where y ∝ 1/x^2, with given values x = 4 and y = 2. This problem illustrates key algebraic concepts in inverse proportionality, suitable for students in Grades 8-10.

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