The distance of the point `(6,-2 sqrt 2)` from the common tangent `y = mx + c, m gt 0`, of the curve `x=2y^2` and `x = 1 + y^2` is:

Options

A. `5sqrt3`
B. 44999
C. 44986
D. 5

Answer

To find the distance of the point \( (6, -2\sqrt{2}) \) from the common tangent \( y = mx + c \) of the curves \( x = 2y^2 \) and \( x = 1 + y^2 \), we will follow these steps:

### Step 1: Determine the equations of the curves
The curves are given as:
1. \( x = 2y^2 \)
2. \( x = 1 + y^2 \)

### Step 2: Find the slopes of the tangents
For the first curve \( x = 2y^2 \):
- The derivative \( \frac{dx}{dy} = 4y \), so the slope \( m_1 = 4y \).

For the second curve \( x = 1 + y^2 \):
- The derivative \( \frac{dx}{dy} = 2y \), so the slope \( m_2 = 2y \).

### Step 3: Set the slopes equal for common tangents
Since we want a common tangent, we set \( 4y_1 = 2y_2 \) where \( y_1 \) and \( y_2 \) are the points of tangency on the first and second curves respectively. This gives us:
\[ y_2 = 2y_1 \]

### Step 4: Find the equations of the tangents
For the first curve at point \( (2y_1^2, y_1) \):
- The equation of the tangent is given by:
\[ x = 2y_1^2 + 4y_1(y - y_1) \]
Rearranging gives:
\[ x - 2y_1^2 = 4y_1y - 4y_1^2 \]
\[ 4y_1y = x - 2y_1^2 + 4y_1^2 \]
\[ y = \frac{x}{4y_1} + \frac{2y_1}{4} \]

For the second curve at point \( (1 + y_2^2, y_2) \):
- The equation of the tangent is:
\[ x = 1 + y_2^2 + 2y_2(y - y_2) \]
Rearranging gives:
\[ x - (1 + y_2^2) = 2y_2(y - y_2) \]
\[ 2y_2y = x - 1 - y_2^2 + 2y_2^2 \]
\[ y = \frac{x - 1}{2y_2} + \frac{y_2}{2} \]

### Step 5: Find the common tangent condition
Equating the two tangent equations, we can find \( y_1 \) and \( y_2 \) in terms of \( m \):
- From the common tangent condition, we derive that:
\[ m = \frac{1}{2\sqrt{2}} \]

### Step 6: Substitute \( m \) into the tangent equation
Now substituting \( m \) back into the tangent equation \( y = mx + c \):
\[ y = \frac{1}{2\sqrt{2}}x + c \]

### Step 7: Find the value of \( c \)
To find \( c \), we can use the condition that the tangent touches both curves. After some calculations, we find:
\[ c = \frac{1}{8m} = \frac{1}{8 \cdot \frac{1}{2\sqrt{2}}} = \frac{\sqrt{2}}{4} \]

### Step 8: Write the final tangent equation
Thus, the equation of the common tangent is:
\[ y = \frac{1}{2\sqrt{2}}x + \frac{\sqrt{2}}{4} \]

### Step 9: Calculate the distance from the point to the tangent
The distance \( d \) from the point \( (6, -2\sqrt{2}) \) to the line \( Ax + By + C = 0 \) can be calculated using the formula:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
Where:
- \( A = -\frac{1}{2\sqrt{2}} \)
- \( B = 1 \)
- \( C = -\frac{\sqrt{2}}{4} \)
- \( (x_0, y_0) = (6, -2\sqrt{2}) \)

Substituting these values:
\[ d = \frac{\left| -\frac{1}{2\sqrt{2}} \cdot 6 + 1 \cdot (-2\sqrt{2}) - \frac{\sqrt{2}}{4} \right|}{\sqrt{\left(-\frac{1}{2\sqrt{2}}\right)^2 + 1^2}} \]

### Step 10: Simplify the expression
Calculating the numerator:
\[ = \left| -\frac{3}{\sqrt{2}} - 2\sqrt{2} - \frac{\sqrt{2}}{4} \right| \]
Combine like terms and simplify.

The denominator simplifies to:
\[ \sqrt{\frac{1}{8} + 1} = \sqrt{\frac{9}{8}} = \frac{3}{2\sqrt{2}} \]

Finally, after performing all calculations, we find that the distance \( d \) is equal to \( 5 \).

### Final Answer
The distance of the point \( (6, -2\sqrt{2}) \) from the common tangent is \( 5 \).

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