The point (0,5) is closest to the curve `x^2 = 2y` at

Options

A. `(2sqrt2,0)`
B. (0,0)
C. `(2,2)`
D. None of these

Answer

To find the point on the curve \( x^2 = 2y \) that is closest to the point \( (0, 5) \), we can follow these steps:

### Step 1: Parametrize the Curve
The curve \( x^2 = 2y \) can be parametrized. We can express points on the curve in terms of a parameter \( t \):
\[ x = t, \quad y = \frac{t^2}{2} \]
Thus, a point on the curve can be represented as \( (t, \frac{t^2}{2}) \).

### Step 2: Distance Formula
Next, we need to find the distance \( D \) between the point \( (0, 5) \) and the point \( (t, \frac{t^2}{2}) \):
\[ D = \sqrt{(t - 0)^2 + \left(\frac{t^2}{2} - 5\right)^2} \]
To simplify our calculations, we will minimize \( D^2 \) instead of \( D \):
\[ D^2 = t^2 + \left(\frac{t^2}{2} - 5\right)^2 \]

### Step 3: Expand the Distance Squared
Now, we expand \( D^2 \):
\[ D^2 = t^2 + \left(\frac{t^2}{2} - 5\right)^2 = t^2 + \left(\frac{t^4}{4} - 5t^2 + 25\right) \]
Combining the terms gives:
\[ D^2 = \frac{t^4}{4} - 4t^2 + 25 \]

### Step 4: Differentiate to Find Critical Points
To find the minimum distance, we differentiate \( D^2 \) with respect to \( t \) and set the derivative to zero:
\[ \frac{d(D^2)}{dt} = t^3 - 8t = 0 \]
Factoring out \( t \):
\[ t(t^2 - 8) = 0 \]
This gives us:
\[ t = 0 \quad \text{or} \quad t^2 = 8 \quad \Rightarrow \quad t = \pm 2\sqrt{2} \]

### Step 5: Second Derivative Test
To determine whether these critical points correspond to a minimum or maximum, we compute the second derivative:
\[ \frac{d^2(D^2)}{dt^2} = 3t^2 - 8 \]
Evaluating at the critical points:
1. For \( t = 0 \):
\[ \frac{d^2(D^2)}{dt^2} = 3(0)^2 - 8 = -8 \quad (\text{maximum}) \]
2. For \( t = 2\sqrt{2} \):
\[ \frac{d^2(D^2)}{dt^2} = 3(2\sqrt{2})^2 - 8 = 24 - 8 = 16 \quad (\text{minimum}) \]
3. For \( t = -2\sqrt{2} \):
\[ \frac{d^2(D^2)}{dt^2} = 3(-2\sqrt{2})^2 - 8 = 24 - 8 = 16 \quad (\text{minimum}) \]

### Step 6: Find the Points on the Curve
Now, we find the corresponding points on the curve:
1. For \( t = 2\sqrt{2} \):
\[ y = \frac{(2\sqrt{2})^2}{2} = \frac{8}{2} = 4 \quad \Rightarrow \quad (2\sqrt{2}, 4) \]
2. For \( t = -2\sqrt{2} \):
\[ y = \frac{(-2\sqrt{2})^2}{2} = \frac{8}{2} = 4 \quad \Rightarrow \quad (-2\sqrt{2}, 4) \]

### Conclusion
The points on the curve \( x^2 = 2y \) that are closest to the point \( (0, 5) \) are:
\[ (2\sqrt{2}, 4) \quad \text{and} \quad (-2\sqrt{2}, 4) \]