Find the locus of a point such that the angle between the tangents from it to the hyperbola `x^(2)/a^(2) - y^(2)/b^(2) = 1` is equal to the angle between the asymptotes of the hyperbola .

Answer

To find the locus of a point such that the angle between the tangents from it to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is equal to the angle between the asymptotes of the hyperbola, we can follow these steps:

### Step 1: Identify the asymptotes of the hyperbola
The asymptotes of the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) are given by the equations:
\[ y = \pm \frac{b}{a} x \]
The angle \( \theta \) between the asymptotes can be calculated using the formula:
\[ \theta = 2 \tan^{-1}\left(\frac{b}{a}\right) \]

### Step 2: Set up the point from which tangents are drawn
Let \( P(h, k) \) be the point from which we are drawing tangents to the hyperbola. The equation of the tangents from point \( P(h, k) \) to the hyperbola can be written as:
\[ \frac{hh_1}{a^2} - \frac{kk_1}{b^2} = 1 \]
where \( (h_1, k_1) \) is a point on the hyperbola.

### Step 3: Find the angle between the tangents
The angle \( \phi \) between the tangents drawn from point \( P(h, k) \) to the hyperbola can be calculated using the formula:
\[ \tan\left(\frac{\phi}{2}\right) = \sqrt{\frac{(h^2/a^2) - 1}{(k^2/b^2)}} \]
Thus, the angle \( \phi \) can be expressed as:
\[ \phi = 2 \tan^{-1}\left(\sqrt{\frac{(h^2/a^2) - 1}{(k^2/b^2)}}\right) \]

### Step 4: Set the angles equal
According to the problem, the angle between the tangents \( \phi \) is equal to the angle between the asymptotes \( \theta \):
\[ 2 \tan^{-1}\left(\sqrt{\frac{(h^2/a^2) - 1}{(k^2/b^2)}}\right) = 2 \tan^{-1}\left(\frac{b}{a}\right) \]

### Step 5: Simplify the equation
Dividing both sides by 2, we have:
\[ \tan^{-1}\left(\sqrt{\frac{(h^2/a^2) - 1}{(k^2/b^2)}}\right) = \tan^{-1}\left(\frac{b}{a}\right) \]
Taking the tangent of both sides gives:
\[ \sqrt{\frac{(h^2/a^2) - 1}{(k^2/b^2)}} = \frac{b}{a} \]

### Step 6: Square both sides
Squaring both sides results in:
\[ \frac{(h^2/a^2) - 1}{(k^2/b^2)} = \frac{b^2}{a^2} \]

### Step 7: Cross-multiply and simplify
Cross-multiplying gives:
\[ (h^2/a^2) - 1 = \frac{b^2 k^2}{a^2} \]
Rearranging leads to:
\[ h^2 = k^2 \frac{b^2}{a^2} + a^2 \]

### Step 8: Form the locus equation
This can be rearranged to form the locus equation:
\[ \frac{h^2}{a^2 + \frac{b^2 k^2}{a^2}} = 1 \]
This represents the locus of the point \( P(h, k) \).

### Conclusion
The locus of the point \( P(h, k) \) is given by the equation derived above.