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To solve this problem, let’s go through it step-by-step.

1. **Define Variables and Relationships:**
   - Let \( P = (x, y) \) be a point on the curve.
   - The equation of the curve is \( y = 4x \), so \( y \) depends on \( x \).
   - We need to find the rate at which the distance between \( P \) and the fixed point \( (2, 0) \) is changing when \( P \) is at \( (2, 8) \) and \( x \) is decreasing at \( 3 \) units/second.

2. **Distance Formula:**
   The distance \( D \) between \( P = (x, y) \) and \( (2, 0) \) is given by:
   \[
   D = \sqrt{(x - 2)^2 + (y - 0)^2} = \sqrt{(x - 2)^2 + y^2}
   \]
   Since \( y = 4x \), substitute \( y = 4x \) into the distance formula:
   \[
   D = \sqrt{(x - 2)^2 + (4x)^2} = \sqrt{(x - 2)^2 + 16x^2}
   \]

3. **Differentiate with Respect to Time \( t \):**
   We want to find \( \frac{dD}{dt} \), the rate at which \( D \) is changing. Differentiate both sides of the distance formula with respect to \( t \):
   \[
   \frac{dD}{dt} = \frac{1}{2} ( (x - 2)^2 + 16x^2 )^{-\frac{1}{2}} \cdot \left( 2(x - 2) \frac{dx}{dt} + 32x \frac{dx}{dt} \right)
   \]
   Simplifying, we get:
   \[
   \frac{dD}{dt} = \frac{(x - 2) \frac{dx}{dt} + 16x \frac{dx}{dt}}{\sqrt{(x - 2)^2 + 16x^2}}
   \]

4. **Substitute Known Values:**
   At the instant we’re considering:
   - \( x = 2 \)
   - \( y = 8 \) (since \( y = 4x \))
   - \( \frac{dx}{dt} = -3 \) units/s

   Substitute these values into the formula:
   \[
   \frac{dD}{dt} = \frac{(2 - 2)(-3) + 16 \cdot 2 \cdot (-3)}{\sqrt{(2 - 2)^2 + 16 \cdot (2)^2}}
   \]
   Simplify each part:
   - The numerator: \( (0)(-3) + 16 \cdot 2 \cdot (-3) = -96 \)
   - The denominator: \( \sqrt{0 + 16 \cdot 4} = \sqrt{64} = 8 \)

   So,
   \[
   \frac{dD}{dt} = \frac{-96}{8} = -12
   \]

5. **Conclusion:**
   The rate at which the distance between \( P \) and \( (2, 0) \) is changing at that instant is \( -12 \) units/second. The negative sign indicates that the distance is decreasing.

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### Related Questions

1. What is the importance of implicit differentiation in related rates problems?
2. How would the solution change if \( x \) were increasing instead of decreasing?
3. Why does the negative rate signify a decreasing distance?
4. What other scenarios might involve a point moving along a curve with changing distances?
5. How can we use related rates in physics problems involving speed and distance?

### Tip
When solving related rates problems, always start by identifying the relationships between the quantities involved before differentiating. This helps prevent errors in differentiation steps.

A point P is moving along the curve whose equation is y = 4x. How fast is the distance between P and the point (2,0) changing at the instant when P is at (2,8) if x is decreasing at the rate of 3 units/s at that instant?

Learn how to calculate the rate at which the distance changes between a point moving along the curve y = 4x and a fixed point (2,0) when x is decreasing at 3 units/s. This problem involves related rates, differentiation, and the distance formula, making it suitable for students in Grades 11-12.

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