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 \( t_1 \) and \( t_2 \) to the hyperbola \( xy = c^2 \), we can follow these steps:

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

### Step 2: Write the Equations of the Tangents
The equations of the tangents to the hyperbola \( xy = c^2 \) at points \( P_{t_1} \) and \( P_{t_2} \) are given by:
- For \( t_1 \):
\[ x + y t_1^2 = 2ct_1 \]
- For \( t_2 \):
\[ x + y t_2^2 = 2ct_2 \]

### Step 3: Subtract the Two Equations
Now, we subtract the second tangent equation from the first:
\[ (x + yt_1^2) - (x + yt_2^2) = 2ct_1 - 2ct_2 \]
This simplifies to:
\[ y(t_1^2 - t_2^2) = 2c(t_1 - t_2) \]

### Step 4: Solve for \( y \)
We can solve for \( y \):
\[ y = \frac{2c(t_1 - t_2)}{t_1^2 - t_2^2} \]
Using the difference of squares, we can rewrite \( t_1^2 - t_2^2 \) as \( (t_1 - t_2)(t_1 + t_2) \):
\[ y = \frac{2c}{t_1 + t_2} \]

### Step 5: Substitute \( y \) Back to Find \( x \)
Now we substitute \( y \) back into one of the tangent equations to find \( x \). Using the tangent equation at \( t_1 \):
\[ x + \left(\frac{2c}{t_1 + t_2}\right)t_1^2 = 2ct_1 \]
Rearranging gives:
\[ x = 2ct_1 - \frac{2ct_1^2}{t_1 + t_2} \]
Taking the common denominator:
\[ x = \frac{2ct_1(t_1 + t_2) - 2ct_1^2}{t_1 + t_2} \]
This simplifies to:
\[ x = \frac{2ct_1t_2}{t_1 + t_2} \]

### Step 6: Final Result
Thus, the point of intersection of the tangents at \( t_1 \) and \( t_2 \) is:
\[ \left( \frac{2ct_1t_2}{t_1 + t_2}, \frac{2c}{t_1 + t_2} \right) \]