Find the locus of the points of intersection of two tangents to a hyperbola `x^(2)/ 25 - y^(2)/ 16 = 1` if sum of their slope is constant a .

Answer

To find the locus of the points of intersection of two tangents to the hyperbola \( \frac{x^2}{25} - \frac{y^2}{16} = 1 \) when the sum of their slopes is a constant \( a \), we can follow these steps:

### Step 1: Identify the hyperbola parameters
The given hyperbola is in the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), where \( a^2 = 25 \) and \( b^2 = 16 \). Thus, \( a = 5 \) and \( b = 4 \).

### Step 2: Write the equation of the tangent
The equation of the tangent to the hyperbola at a point \( (x_0, y_0) \) is given by:
\[ y = mx \pm \sqrt{a^2 m^2 - b^2} \]
For our hyperbola, this becomes:
\[ y = mx \pm \sqrt{25m^2 - 16} \]

### Step 3: Set up the point of intersection
Let \( P(h, k) \) be the point of intersection of the two tangents. The slopes of the two tangents are \( m_1 \) and \( m_2 \), and we know that:
\[ m_1 + m_2 = a \quad \text{(constant)} \]

### Step 4: Express \( k \) in terms of \( h \) and \( m \)
From the equation of the tangent, we can express \( k \) as:
\[ k = m h \pm \sqrt{25m^2 - 16} \]
This leads to two equations for \( k \):
1. \( k = mh + \sqrt{25m^2 - 16} \)
2. \( k = mh - \sqrt{25m^2 - 16} \)

### Step 5: Square the equations
We can square both sides of the equations to eliminate the square root:
\[ (k - mh)^2 = 25m^2 - 16 \]
Expanding this gives:
\[ k^2 - 2mhk + m^2h^2 = 25m^2 - 16 \]

### Step 6: Rearranging the equation
Rearranging the equation leads to:
\[ k^2 - 2mhk + m^2h^2 - 25m^2 + 16 = 0 \]
This is a quadratic equation in \( m \).

### Step 7: Use the quadratic formula
For a quadratic equation \( Am^2 + Bm + C = 0 \), the sum of the roots \( m_1 + m_2 \) is given by \( -\frac{B}{A} \). Here:
- \( A = h^2 - 25 \)
- \( B = -2hk \)
- \( C = k^2 + 16 \)

Thus:
\[ m_1 + m_2 = \frac{2hk}{h^2 - 25} \]

### Step 8: Set the sum of slopes equal to \( a \)
Since \( m_1 + m_2 = a \), we have:
\[ \frac{2hk}{h^2 - 25} = a \]

### Step 9: Rearranging to find the locus
Multiplying through by \( h^2 - 25 \) gives:
\[ 2hk = a(h^2 - 25) \]
Rearranging this results in:
\[ ah^2 - 2hk - 25a = 0 \]

### Step 10: Replace \( h \) and \( k \) with \( x \) and \( y \)
Finally, replacing \( h \) with \( x \) and \( k \) with \( y \), we get the locus:
\[ ax^2 - 2xy - 25a = 0 \]

### Final Answer
The locus of the points of intersection of the two tangents is:
\[ ax^2 - 2xy - 25a = 0 \]