The maximum distance of the point `(k, 0)` from the curve `2x^2+y^2-2x=0` is equal to

Options

A. `sqrt(1+2k-k^(2))`
B. `sqrt(1+2k+2k^(2))`
C. `sqrt(1-2k+2k^(2))`
D. `sqrt(1-2k+k^(2))`

Answer

To solve the problem of finding the maximum distance from the point \((k, 0)\) to the curve given by the equation \(2x^2 + y^2 - 2x = 0\), we will follow these steps:

### Step 1: Rearranging the Curve Equation
We start with the equation of the curve:
\[ 2x^2 + y^2 - 2x = 0 \]
Rearranging this gives us:
\[ y^2 = 2x - 2x^2 \]

### Step 2: Finding the Points on the Curve
To find points on the curve, we can express \(y\) in terms of \(x\):
\[ y = \sqrt{2x - 2x^2} \]
Let \(P\) be a point on the curve, where \(P = (\alpha, \sqrt{2\alpha - 2\alpha^2})\).

### Step 3: Distance Formula
The distance \(d\) between the point \((k, 0)\) and the point \(P\) is given by:
\[ d^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2 \]
Substituting the coordinates:
\[ d^2 = (\alpha - k)^2 + \left(\sqrt{2\alpha - 2\alpha^2} - 0\right)^2 \]
This simplifies to:
\[ d^2 = (\alpha - k)^2 + (2\alpha - 2\alpha^2) \]

### Step 4: Expanding the Distance Formula
Now we expand \(d^2\):
\[ d^2 = (\alpha - k)^2 + 2\alpha - 2\alpha^2 \]
Expanding \((\alpha - k)^2\):
\[ d^2 = \alpha^2 - 2k\alpha + k^2 + 2\alpha - 2\alpha^2 \]
Combining like terms gives:
\[ d^2 = -\alpha^2 - 2k\alpha + k^2 + 2\alpha \]
This can be rewritten as:
\[ d^2 = -\alpha^2 + (2 - 2k)\alpha + k^2 \]

### Step 5: Finding the Maximum Distance
To find the maximum distance, we need to differentiate \(d^2\) with respect to \(\alpha\) and set it to zero:
\[ \frac{d(d^2)}{d\alpha} = -2\alpha + (2 - 2k) = 0 \]
Solving for \(\alpha\):
\[ -2\alpha + 2 - 2k = 0 \implies 2\alpha = 2 - 2k \implies \alpha = 1 - k \]

### Step 6: Substituting Back to Find \(d^2\)
Now we substitute \(\alpha = 1 - k\) back into the distance formula:
\[ d^2 = -(1 - k)^2 + (2 - 2k)(1 - k) + k^2 \]
Calculating each term:
1. \(-(1 - k)^2 = -1 + 2k - k^2\)
2. \((2 - 2k)(1 - k) = 2 - 2k - 2 + 2k^2 = 2k^2 - 2k\)
3. \(k^2\)

Combining these:
\[ d^2 = (-1 + 2k - k^2) + (2k^2 - 2k) + k^2 \]
Simplifying:
\[ d^2 = -1 + 2k - k^2 + 2k^2 - 2k + k^2 = -1 + 2k^2 \]
Thus:
\[ d^2 = 1 + 2k^2 - 2k \]

### Step 7: Final Result for Distance
The maximum distance \(d\) is:
\[ d = \sqrt{1 + 2k^2 - 2k} \]