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 from which two tangents are drawn to the hyperbola given by the equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), under the condition that the product of the slopes of these tangents is equal to 1.

### Step 1: Define the point P
Let the coordinates of point P be \( (h, k) \).

### Step 2: Write the equation of the tangents
The equation of the tangents drawn from the point \( (h, k) \) to the hyperbola can be expressed using the formula:
\[ k = mh \pm \sqrt{a^2 m^2 - b^2} \]
where \( m \) is the slope of the tangent.

### Step 3: Rearrange the equation
Rearranging the equation gives:
\[ k - mh = \pm \sqrt{a^2 m^2 - b^2} \]
Squaring both sides results in:
\[ (k - mh)^2 = a^2 m^2 - b^2 \]
Expanding this, we have:
\[ k^2 - 2kmh + m^2h^2 = a^2 m^2 - b^2 \]
Rearranging gives:
\[ m^2(h^2 - a^2) - 2kmh + (k^2 + b^2) = 0 \]

### Step 4: Identify the coefficients
This is a quadratic equation in \( m \):
\[ Am^2 + Bm + C = 0 \]
where:
- \( A = h^2 - a^2 \)
- \( B = -2kh \)
- \( C = k^2 + b^2 \)

### Step 5: Use the condition on the product of slopes
The product of the slopes of the tangents is given by the formula:
\[ \text{Product of roots} = \frac{C}{A} = 1 \]
Substituting for \( C \) and \( A \):
\[ \frac{k^2 + b^2}{h^2 - a^2} = 1 \]
This implies:
\[ k^2 + b^2 = h^2 - a^2 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ h^2 - k^2 = a^2 + b^2 \]

### Step 7: Recognize the conic section
The equation \( h^2 - k^2 = a^2 + b^2 \) represents a hyperbola in the \( hk \)-plane.

### Step 8: Find the eccentricity
The standard form of a hyperbola is given by:
\[ \frac{h^2}{a^2 + b^2} - \frac{k^2}{a^2 + b^2} = 1 \]
The eccentricity \( e \) of a hyperbola is given by:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
In our case, since we have \( a^2 + b^2 \) in the denominator, we can express the eccentricity as:
\[ e' = \sqrt{1 + \frac{a^2 + b^2}{a^2}} = \sqrt{1 + \frac{b^2}{a^2 + b^2}} = \sqrt{2} \]

Thus, the eccentricity of the conic whose locus is defined by the given conditions is \( \sqrt{2} \).

### Final Answer:
The eccentricity is \( \sqrt{2} \).