From a point P, two tangents PA and PB are drawn to the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`. If the product of the slopes of these tangents is 1, then the locus of P is a conic whose eccentricity is equal to

Options

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

Answer

To solve the problem step by step, we need to find the locus of the point \( P(h, k) \) from which two tangents \( PA \) and \( PB \) are drawn to the hyperbola given by the equation:

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

### Step 1: Write the equation of the tangent to the hyperbola
The equation of the tangent to the hyperbola at point \( (x_0, y_0) \) can be expressed in slope form as:

\[ y - mx = \pm \sqrt{a^2 m^2 - b^2} \]

where \( m \) is the slope of the tangent.

### Step 2: Substitute the coordinates of point \( P(h, k) \)
Since the tangents \( PA \) and \( PB \) are drawn from point \( P(h, k) \), we substitute \( x = h \) and \( y = k \) into the tangent equation:

\[ k - mh = \pm \sqrt{a^2 m^2 - b^2} \]

### Step 3: Square both sides
To eliminate the square root, we square both sides:

\[ (k - mh)^2 = a^2 m^2 - b^2 \]

Expanding the left side gives:

\[ k^2 - 2kmh + m^2h^2 = a^2 m^2 - b^2 \]

### Step 4: Rearrange the equation
Rearranging the equation, we get:

\[ (k^2 + b^2) = (a^2 - h^2)m^2 + 2kmh \]

This is a quadratic equation in \( m \):

\[ (a^2 - h^2)m^2 + 2kmh - (k^2 + b^2) = 0 \]

### Step 5: Condition for tangents
For the point \( P(h, k) \) to have two tangents, the discriminant of this quadratic must be zero:

\[ D = (2kh)^2 - 4(a^2 - h^2)(-(k^2 + b^2)) = 0 \]

### Step 6: Simplify the discriminant
Simplifying the discriminant gives:

\[ 4k^2h^2 + 4(k^2 + b^2)(a^2 - h^2) = 0 \]

### Step 7: Rearranging
This leads to:

\[ k^2h^2 + (k^2 + b^2)(a^2 - h^2) = 0 \]

### Step 8: Further simplification
Expanding and rearranging gives us:

\[ k^2a^2 - k^2h^2 + b^2a^2 - b^2h^2 = 0 \]

### Step 9: Expressing in terms of \( h \) and \( k \)
This can be rearranged to find the relationship between \( h \) and \( k \):

\[ k^2(a^2 + b^2) = h^2(a^2 + b^2) \]

### Step 10: Final equation
Thus, we arrive at:

\[ k^2 = h^2 \]

This implies:

\[ \frac{h^2}{k^2} = 1 \]

### Conclusion: Identify the conic
The equation \( h^2 - k^2 = 0 \) represents a rectangular hyperbola, and the eccentricity \( e \) of a rectangular hyperbola is given by:

\[ e = \sqrt{2} \]

### Final Answer
Thus, the eccentricity of the conic whose locus is defined by the given condition is:

\[ \sqrt{2} \]