The locus of a point `P(alpha, beta)` moving under the condtion that the line `y=alphax+beta` is a tangent to the hyperbola `(x^(2))/(a^2)-(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 locus of the point \( P(\alpha, \beta) \) such that the line \( y = \alpha x + \beta \) is a tangent to the hyperbola given by

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

### Step-by-Step Solution:

1. **Identify the Tangent Line Equation**:
The equation of the line is given as \( y = \alpha x + \beta \). This line will be tangent to the hyperbola.

2. **General Tangent Equation for Hyperbola**:
The general form of the tangent to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) at the point \( (x_0, y_0) \) is given by:
\[ \frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1. \]
If we let the slope of the tangent line be \( m \), the tangent line can also be expressed as:
\[ y = mx + \sqrt{a^2m^2 - b^2}. \]

3. **Comparing the Two Equations**:
By comparing the two equations \( y = \alpha x + \beta \) and \( y = mx + \sqrt{a^2m^2 - b^2} \), we can equate the coefficients:
- From the slopes, we have \( \alpha = m \).
- From the constant terms, we have \( \beta = \sqrt{a^2m^2 - b^2} \).

4. **Substituting for \( m \)**:
Substituting \( m = \alpha \) into the equation for \( \beta \):
\[ \beta = \sqrt{a^2\alpha^2 - b^2}. \]

5. **Squaring Both Sides**:
To eliminate the square root, we square both sides:
\[ \beta^2 = a^2\alpha^2 - b^2. \]

6. **Rearranging the Equation**:
Rearranging gives us:
\[ a^2\alpha^2 - \beta^2 = b^2. \]

7. **Substituting \( \alpha \) and \( \beta \) with \( x \) and \( y \)**:
To find the locus, we replace \( \alpha \) with \( x \) and \( \beta \) with \( y \):
\[ a^2x^2 - y^2 = b^2. \]

8. **Rearranging to Standard Form**:
Rearranging gives:
\[ \frac{x^2}{b^2} - \frac{y^2}{a^2} = 1. \]
This is the equation of a hyperbola.

9. **Finding the Eccentricity**:
The eccentricity \( e \) of a hyperbola is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}}. \]
Since we have \( a^2 = b^2 \) (from the hyperbola's equation), we substitute:
\[ e = \sqrt{1 + \frac{b^2}{b^2}} = \sqrt{1 + 1} = \sqrt{2}. \]

### Conclusion:
Thus, the eccentricity of the locus of the point \( P(\alpha, \beta) \) is \( \sqrt{2} \).