The points on the curve `y=x^(2)` which are closest to the point P(0, 1) are

Options

A. `(pm sqrt2,2)`
B. `(pm(1)/(2),(1)/(4))`
C. `(pm(1)/(sqrt2),(1)/(2))`
D. `(pm(1)/(4),(1)/(16))`

Answer

To find the points on the curve \( y = x^2 \) that are closest to the point \( P(0, 1) \), we will follow these steps:

### Step 1: Define the distance function
The distance \( D \) between a point \( Q(t, t^2) \) on the curve and the point \( P(0, 1) \) can be expressed using the distance formula:
\[ D = \sqrt{(t - 0)^2 + (t^2 - 1)^2} \]
This simplifies to:
\[ D = \sqrt{t^2 + (t^2 - 1)^2} \]

### Step 2: Simplify the distance function
We can simplify the expression inside the square root:
\[ D = \sqrt{t^2 + (t^4 - 2t^2 + 1)} = \sqrt{t^4 - t^2 + 1} \]

### Step 3: Minimize the distance function
To minimize \( D \), we can minimize \( D^2 \) (since the square root function is monotonically increasing):
\[ f(t) = t^4 - t^2 + 1 \]

### Step 4: Find the derivative
Next, we find the derivative of \( f(t) \):
\[ f'(t) = 4t^3 - 2t \]

### Step 5: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:
\[ 4t^3 - 2t = 0 \]
Factoring out \( 2t \):
\[ 2t(2t^2 - 1) = 0 \]
This gives us:
\[ t = 0 \quad \text{or} \quad 2t^2 - 1 = 0 \implies t^2 = \frac{1}{2} \implies t = \pm \frac{1}{\sqrt{2}} \]

### Step 6: Determine the nature of critical points
We need to check the second derivative to determine if these points are minima:
\[ f''(t) = 12t^2 - 2 \]
- For \( t = 0 \):
\[ f''(0) = -2 \quad (\text{not a minimum}) \]
- For \( t = \frac{1}{\sqrt{2}} \):
\[ f''\left(\frac{1}{\sqrt{2}}\right) = 12 \cdot \frac{1}{2} - 2 = 6 \quad (\text{minimum}) \]
- For \( t = -\frac{1}{\sqrt{2}} \):
\[ f''\left(-\frac{1}{\sqrt{2}}\right) = 12 \cdot \frac{1}{2} - 2 = 6 \quad (\text{minimum}) \]

### Step 7: Find the points on the curve
Now we can find the corresponding points on the curve \( y = x^2 \):
- For \( t = \frac{1}{\sqrt{2}} \):
\[ Q\left(\frac{1}{\sqrt{2}}, \left(\frac{1}{\sqrt{2}}\right)^2\right) = \left(\frac{1}{\sqrt{2}}, \frac{1}{2}\right) \]
- For \( t = -\frac{1}{\sqrt{2}} \):
\[ Q\left(-\frac{1}{\sqrt{2}}, \left(-\frac{1}{\sqrt{2}}\right)^2\right) = \left(-\frac{1}{\sqrt{2}}, \frac{1}{2}\right) \]

### Final Answer
The points on the curve \( y = x^2 \) that are closest to the point \( P(0, 1) \) are:
\[ \left(\frac{1}{\sqrt{2}}, \frac{1}{2}\right) \quad \text{and} \quad \left(-\frac{1}{\sqrt{2}}, \frac{1}{2}\right) \]