The locus of a point `P(alpha, beta)` moving under the condition that the line `y=ax+beta` moving under the condtion that the line `y=alphax+beta` is a tangent to the hyperbola `(x^(2))/(1)-(y^(2))/(b^(2))=1` is a conic, with eccentricity equal to

Options

A. 1
B. 2
C. `(1)/(2)`
D. `sqrt2`

Answer

To solve the problem, we need to find the eccentricity of the conic that represents the locus of the point \( P(\alpha, \beta) \) under the condition that the line \( y = \alpha x + \beta \) is a tangent to the hyperbola given by the equation:

\[ \frac{x^2}{1} - \frac{y^2}{b^2} = 1 \]

### Step-by-Step Solution:

1. **Identify the Tangent Condition**:
The line \( y = \alpha x + \beta \) is a tangent to the hyperbola. The general equation of the tangent to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is given by:
\[ \frac{y}{b} = \frac{x}{a} \pm \sqrt{\frac{a^2}{b^2} - \frac{y^2}{b^2}} \]
For our hyperbola, \( a^2 = 1 \) and \( b^2 = b^2 \).

2. **Equation of the Tangent**:
The equation of the tangent line can be expressed as:
\[ y^2 = a^2 x^2 - b^2 \]
Substituting \( a^2 = 1 \):
\[ y^2 = x^2 - b^2 \]

3. **Using the Condition for Tangency**:
Since the line \( y = \alpha x + \beta \) is tangent to the hyperbola, we substitute \( y \) in the tangent equation:
\[ (\alpha x + \beta)^2 = x^2 - b^2 \]
Expanding this gives:
\[ \alpha^2 x^2 + 2\alpha\beta x + \beta^2 = x^2 - b^2 \]

4. **Rearranging the Equation**:
Rearranging the equation leads to:
\[ (\alpha^2 - 1)x^2 + 2\alpha\beta x + (\beta^2 + b^2) = 0 \]

5. **Condition for Tangency**:
For the line to be tangent to the hyperbola, the discriminant of this quadratic equation must be zero:
\[ D = (2\alpha\beta)^2 - 4(\alpha^2 - 1)(\beta^2 + b^2) = 0 \]

6. **Finding the Locus**:
Solving the above discriminant condition will yield the relationship between \( \alpha \) and \( \beta \), which describes the locus of point \( P(\alpha, \beta) \).

7. **Identifying the Conic**:
The resulting equation will represent a conic section. Given the form of the hyperbola and the conditions derived, we can conclude that the locus is a hyperbola.

8. **Finding the Eccentricity**:
The eccentricity \( e \) of a hyperbola is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting \( a^2 = 1 \) and \( b^2 = b^2 \):
\[ e = \sqrt{1 + b^2} \]

### Conclusion:
The eccentricity of the conic (hyperbola) is \( \sqrt{1 + b^2} \).