If the line `(x-1)/(5)=(y-3)/(2)=(z-3)/(2)` intersects the curve `x^(2)-y^(2)=k^(2), z=0`, then the value of 2k can be equal to

Options

A. `-13`
B. `-(13)/(2)`
C. `(13)/(2)`
D. `(13)/(4)`

Answer

To solve the problem step by step, we will analyze the given line and the curve, and find the intersection points.

### Step 1: Understand the line equation
The line is given in the symmetric form:
\[ \frac{x-1}{5} = \frac{y-3}{2} = \frac{z-3}{2} \]
We can denote this common ratio as \( t \). Therefore, we can express \( x, y, z \) in terms of \( t \):
\[ x = 5t + 1, \quad y = 2t + 3, \quad z = 2t + 3 \]

### Step 2: Set \( z = 0 \)
Since we are interested in the intersection with the plane \( z = 0 \), we set \( z = 0 \):
\[ 2t + 3 = 0 \implies 2t = -3 \implies t = -\frac{3}{2} \]

### Step 3: Substitute \( t \) back to find \( x \) and \( y \)
Now we substitute \( t = -\frac{3}{2} \) back into the equations for \( x \) and \( y \):
\[ x = 5\left(-\frac{3}{2}\right) + 1 = -\frac{15}{2} + 1 = -\frac{15}{2} + \frac{2}{2} = -\frac{13}{2} \]
\[ y = 2\left(-\frac{3}{2}\right) + 3 = -3 + 3 = 0 \]

### Step 4: Substitute \( x \) and \( y \) into the curve equation
The curve is given by:
\[ x^2 - y^2 = k^2 \]
Substituting \( x = -\frac{13}{2} \) and \( y = 0 \):
\[ \left(-\frac{13}{2}\right)^2 - 0^2 = k^2 \]
\[ \frac{169}{4} = k^2 \]

### Step 5: Solve for \( k \)
Taking the square root of both sides, we find:
\[ k = \pm \frac{13}{2} \]

### Step 6: Find \( 2k \)
Now we calculate \( 2k \):
\[ 2k = 2 \times \pm \frac{13}{2} = \pm 13 \]

### Conclusion
Thus, the possible values of \( 2k \) are \( 13 \) and \( -13 \). The answer is \( \pm 13 \).