Consider a hyperbola: `((x-7)^(2))/(a) -((y+3)^(2))/(b^(2)) =1`. The line `3x - 2y - 25 =0`, which is not a tangent, intersect the hyperbola at `H ((11)/(3),-7)` only. A variable point `P(alpha +7, alpha^(2)-4) AA alpha in R` exists in the plane of the given hyperbola.
The eccentricity of the hyperbola is

Options

A. `sqrt((7)/(5))`
B. `sqrt(2)`
C. `(sqrt(13))/(2)`
D. `(3)/(2)`

Explanation

To solve the problem step by step, we will analyze the given hyperbola and the line, and then we will find the eccentricity of the hyperbola.

### Step 1: Identify the equation of the hyperbola
The hyperbola is given by the equation:
\[ \frac{(x-7)^2}{a} - \frac{(y+3)^2}{b^2} = 1 \]
This is in the standard form of a hyperbola centered at (7, -3).

### Step 2: Analyze the line
The line given is:
\[ 3x - 2y - 25 = 0 \]
We can rewrite this in slope-intercept form \(y = mx + c\):
\[ 2y = 3x - 25 \implies y = \frac{3}{2}x - \frac{25}{2} \]
The slope \(m\) of the line is \(\frac{3}{2}\).

### Step 3: Determine the relationship between the line and the hyperbola
Since the line intersects the hyperbola at only one point and is not a tangent, it must be an asymptote of the hyperbola. The slope of the asymptotes of a hyperbola in the form given is \(\pm \frac{b}{a}\).

Thus, we have:
\[ \frac{b}{a} = \frac{3}{2} \]

### Step 4: Express \(b\) in terms of \(a\)
From the relationship \(\frac{b}{a} = \frac{3}{2}\), we can express \(b\) as:
\[ b = \frac{3}{2}a \]

### Step 5: Calculate the eccentricity of the hyperbola
The eccentricity \(e\) of a hyperbola is given by the formula:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting \(b = \frac{3}{2}a\) into the eccentricity formula:
\[ e = \sqrt{1 + \frac{\left(\frac{3}{2}a\right)^2}{a^2}} = \sqrt{1 + \frac{\frac{9}{4}a^2}{a^2}} = \sqrt{1 + \frac{9}{4}} = \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2} \]

### Step 6: Final answer
Thus, the eccentricity of the hyperbola is:
\[ \frac{\sqrt{13}}{2} \]