The point of intersection of tangents at `'t'_(1) " and " 't'_(2)` to the hyperbola `xy = c^(2)` is

Options

A. `((ct_(1)t_(2))/(t_(1) + t_(2)) , c/(t_(1)+t_(2))) `
B. `((2ct_(1)t_(2))/(t_(1) + t_(2)) , (2c)/(t_(1) + t_(2)))`
C. `((t_(1)t_(2))/(c(t_(1) + t_(2))), (t_(1) + t_(2))/c)`
D. none of these

Answer

To find the point of intersection of the tangents at points \( t_1 \) and \( t_2 \) to the hyperbola \( xy = c^2 \), we can follow these steps:

### Step 1: Identify the Points
The points on the hyperbola corresponding to parameters \( t_1 \) and \( t_2 \) can be represented as:
- Point corresponding to \( t_1 \): \( P(t_1) = (ct_1, \frac{c^2}{t_1}) \)
- Point corresponding to \( t_2 \): \( P(t_2) = (ct_2, \frac{c^2}{t_2}) \)

### Step 2: Write the Equation of the Tangents
The equation of the tangent to the hyperbola \( xy = c^2 \) at a point \( (x_0, y_0) \) is given by:
\[ x + y \cdot t^2 = 2ct \]
For point \( P(t_1) \):
\[ x + y t_1^2 = 2ct_1 \quad \text{(Equation 1)} \]
For point \( P(t_2) \):
\[ x + y t_2^2 = 2ct_2 \quad \text{(Equation 2)} \]

### Step 3: Solve the System of Equations
We now have two equations:
1. \( x + yt_1^2 = 2ct_1 \)
2. \( x + yt_2^2 = 2ct_2 \)

To find the point of intersection, we can subtract Equation 1 from Equation 2:
\[ (yt_2^2 - yt_1^2) = 2ct_2 - 2ct_1 \]
This simplifies to:
\[ y(t_2^2 - t_1^2) = 2c(t_2 - t_1) \]
From this, we can solve for \( y \):
\[ y = \frac{2c(t_2 - t_1)}{t_2^2 - t_1^2} \]
Using the difference of squares, we can rewrite \( t_2^2 - t_1^2 \) as \( (t_2 - t_1)(t_2 + t_1) \):
\[ y = \frac{2c}{t_2 + t_1} \]

### Step 4: Substitute \( y \) Back to Find \( x \)
Now, substitute \( y \) back into either Equation 1 or Equation 2 to find \( x \). Using Equation 1:
\[ x + \left(\frac{2c}{t_2 + t_1}\right)t_1^2 = 2ct_1 \]
Rearranging gives:
\[ x = 2ct_1 - \frac{2ct_1^2}{t_2 + t_1} \]
Factoring out \( 2c \):
\[ x = 2c\left(t_1 - \frac{t_1^2}{t_2 + t_1}\right) \]
This simplifies to:
\[ x = \frac{2ct_1 t_2}{t_2 + t_1} \]

### Step 5: Final Coordinates of the Intersection Point
Thus, the coordinates of the point of intersection of the tangents at \( t_1 \) and \( t_2 \) are:
\[ \left(\frac{2ct_1 t_2}{t_1 + t_2}, \frac{2c}{t_1 + t_2}\right) \]

### Summary
The point of intersection of the tangents at \( t_1 \) and \( t_2 \) to the hyperbola \( xy = c^2 \) is:
\[ \left(\frac{2ct_1 t_2}{t_1 + t_2}, \frac{2c}{t_1 + t_2}\right) \]