The point (0,3) is nearest to the curve ` x^(2)=2y` at

Options

A. `(2sqrt(2),0)`
B. `(0,0)`
C. (2,2)
D. (2,-2)

Answer

To find the point on the curve \( x^2 = 2y \) that is nearest to the point \( (0, 3) \), we can follow these steps:

### Step 1: Express the distance
Let \( P(x, y) \) be a point on the curve. The distance \( D \) from the point \( (0, 3) \) to the point \( P(x, y) \) can be expressed as:
\[ D = \sqrt{(x - 0)^2 + (y - 3)^2} = \sqrt{x^2 + (y - 3)^2} \]

### Step 2: Substitute \( y \) in terms of \( x \)
Since the point \( P \) lies on the curve \( x^2 = 2y \), we can express \( y \) in terms of \( x \):
\[ y = \frac{x^2}{2} \]
Now substitute this into the distance formula:
\[ D = \sqrt{x^2 + \left(\frac{x^2}{2} - 3\right)^2} \]

### Step 3: Simplify the distance function
Now, simplify the expression inside the square root:
\[ D = \sqrt{x^2 + \left(\frac{x^2}{2} - 3\right)^2} \]
\[ = \sqrt{x^2 + \left(\frac{x^4}{4} - 3x^2 + 9\right)} \]
\[ = \sqrt{\frac{x^4}{4} - 2x^2 + 9} \]

### Step 4: Minimize the distance
To minimize \( D \), we can minimize \( D^2 \) (since the square root is a monotonically increasing function):
\[ D^2 = \frac{x^4}{4} - 2x^2 + 9 \]
Let \( f(x) = \frac{x^4}{4} - 2x^2 + 9 \).

### Step 5: Differentiate and find critical points
Now differentiate \( f(x) \):
\[ f'(x) = x^3 - 4x \]
Set the derivative equal to zero to find critical points:
\[ x^3 - 4x = 0 \]
Factoring gives:
\[ x(x^2 - 4) = 0 \]
Thus, \( x = 0 \) or \( x^2 - 4 = 0 \) which gives \( x = \pm 2 \).

### Step 6: Evaluate the critical points
Now, we need to evaluate \( y \) at these \( x \) values:
1. For \( x = 0 \):
\[ y = \frac{0^2}{2} = 0 \quad \text{(Point: (0, 0))} \]
2. For \( x = 2 \):
\[ y = \frac{2^2}{2} = 2 \quad \text{(Point: (2, 2))} \]
3. For \( x = -2 \):
\[ y = \frac{(-2)^2}{2} = 2 \quad \text{(Point: (-2, 2))} \]

### Step 7: Determine the nearest point
Now we need to find which of these points is closest to \( (0, 3) \):
- Distance to \( (0, 0) \):
\[ D = \sqrt{(0 - 0)^2 + (0 - 3)^2} = \sqrt{9} = 3 \]
- Distance to \( (2, 2) \):
\[ D = \sqrt{(2 - 0)^2 + (2 - 3)^2} = \sqrt{4 + 1} = \sqrt{5} \]
- Distance to \( (-2, 2) \):
\[ D = \sqrt{(-2 - 0)^2 + (2 - 3)^2} = \sqrt{4 + 1} = \sqrt{5} \]

### Conclusion
The points \( (2, 2) \) and \( (-2, 2) \) are both at a distance of \( \sqrt{5} \) from \( (0, 3) \). Therefore, the nearest points on the curve to \( (0, 3) \) are \( (2, 2) \) and \( (-2, 2) \).

### Final Answer
The point nearest to the curve \( x^2 = 2y \) from \( (0, 3) \) is \( (2, 2) \) or \( (-2, 2) \).

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