The tangent at a point `P` on the hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1` passes through the point `(0,-b)` and the normal at `P` passes through the point `(2asqrt(2),0)` . Then the eccentricity of the hyperbola is

Options

A. a. 2
B. b. `sqrt2`
C. c. 3
D. d. `sqrt3`

Explanation

To solve the problem, we need to find the eccentricity of the hyperbola given the conditions about the tangent and normal lines at a point \( P \) on the hyperbola.

### Step-by-Step Solution:

1. **Identify the Hyperbola**: The hyperbola is given by the equation:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

2. **Assume Point P**: Let the coordinates of point \( P \) on the hyperbola be \( (x_1, y_1) \).

3. **Equation of the Tangent**: The equation of the tangent at point \( P \) is given by:
\[ \frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1 \]
Since the tangent passes through the point \( (0, -b) \), we substitute \( x = 0 \) and \( y = -b \):
\[ \frac{0 \cdot x_1}{a^2} - \frac{(-b)y_1}{b^2} = 1 \]
Simplifying this gives:
\[ \frac{by_1}{b^2} = 1 \implies y_1 = b \]

4. **Coordinates of Point P**: Now we have the coordinates of point \( P \) as \( (x_1, b) \).

5. **Equation of the Normal**: The equation of the normal at point \( P \) is given by:
\[ \frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 e^2 \]
Substituting \( (2a\sqrt{2}, 0) \) into the normal equation:
\[ \frac{a^2 (2a\sqrt{2})}{x_1} + \frac{b^2 \cdot 0}{b} = a^2 e^2 \]
This simplifies to:
\[ \frac{2a^3\sqrt{2}}{x_1} = a^2 e^2 \]
Rearranging gives:
\[ x_1 = \frac{2a^3\sqrt{2}}{a^2 e^2} = \frac{2a\sqrt{2}}{e^2} \]

6. **Substituting into Hyperbola Equation**: Since point \( P \) lies on the hyperbola, we substitute \( (x_1, y_1) = \left(\frac{2a\sqrt{2}}{e^2}, b\right) \) into the hyperbola equation:
\[ \frac{\left(\frac{2a\sqrt{2}}{e^2}\right)^2}{a^2} - \frac{b^2}{b^2} = 1 \]
Simplifying this gives:
\[ \frac{8a^2}{e^4 a^2} - 1 = 1 \]
\[ \frac{8}{e^4} - 1 = 1 \implies \frac{8}{e^4} = 2 \implies e^4 = 4 \]

7. **Finding Eccentricity**: Taking the square root gives:
\[ e^2 = 2 \implies e = \sqrt{2} \]

### Final Answer:
The eccentricity of the hyperbola is:
\[ \boxed{\sqrt{2}} \]