Find the point on the curve `y^(2)=2x`, which is nearest to the point `(1, -4)`.

Answer

To find the point on the curve \( y^2 = 2x \) that is nearest to the point \( (1, -4) \), we can follow these steps:

### Step 1: Set up the distance formula
The distance \( d \) between the point \( (x, y) \) on the curve and the point \( (1, -4) \) is given by the distance formula:
\[ d = \sqrt{(x - 1)^2 + (y + 4)^2} \]

### Step 2: Express \( x \) in terms of \( y \)
From the curve equation \( y^2 = 2x \), we can express \( x \) in terms of \( y \):
\[ x = \frac{y^2}{2} \]

### Step 3: Substitute \( x \) into the distance formula
Substituting \( x \) into the distance formula gives:
\[ d = \sqrt{\left(\frac{y^2}{2} - 1\right)^2 + (y + 4)^2} \]

### Step 4: Minimize the distance
To minimize the distance, we can minimize \( d^2 \) (since the square root function is increasing). Thus, we define:
\[ d^2 = \left(\frac{y^2}{2} - 1\right)^2 + (y + 4)^2 \]

### Step 5: Differentiate \( d^2 \)
Let \( f(y) = \left(\frac{y^2}{2} - 1\right)^2 + (y + 4)^2 \).
Now, we differentiate \( f(y) \):
\[ f'(y) = 2\left(\frac{y^2}{2} - 1\right) \cdot \left(y\right) + 2(y + 4) \]
\[ = y\left(\frac{y^2}{2} - 1\right) + (y + 4) \]

### Step 6: Set the derivative to zero
Setting \( f'(y) = 0 \) for minimization:
\[ y\left(\frac{y^2}{2} - 1\right) + (y + 4) = 0 \]
This simplifies to:
\[ \frac{y^3}{2} - y + y + 4 = 0 \]
\[ \frac{y^3}{2} + 4 = 0 \]
\[ y^3 = -8 \]
\[ y = -2 \]

### Step 7: Find the corresponding \( x \)
Now, substituting \( y = -2 \) back into the equation \( y^2 = 2x \):
\[ (-2)^2 = 2x \implies 4 = 2x \implies x = 2 \]

### Step 8: Conclusion
Thus, the point on the curve that is nearest to the point \( (1, -4) \) is:
\[ \boxed{(2, -2)} \]