If there are two points `A and B` on rectangular hyperbola `xy=c^2` such that abscissa of `A =` ordinate of `B,` then locusof point of intersection of tangents at `A and B` is (a) `y^2-x^2=2c^2` (b) `y^2-x^2=c^2/2` (c) `y=x` (d) non of these

Options

A. `y^(2)=x^(2)+2c^(2)`
B. `y^(2)=x^(2)+(c^(2))/(2)`
C. `y=x`
D. `y=3x`

Answer

To solve the problem, we need to find the locus of the point of intersection of the tangents at points A and B on the rectangular hyperbola defined by the equation \(xy = c^2\). The condition given is that the abscissa (x-coordinate) of point A is equal to the ordinate (y-coordinate) of point B.

### Step-by-Step Solution

1. **Identify Points A and B**:
Let the coordinates of point A be \(A\left(c t_1, \frac{c^2}{t_1}\right)\) and the coordinates of point B be \(B\left(c t_2, \frac{c^2}{t_2}\right)\).

2. **Use the Given Condition**:
According to the problem, the abscissa of A equals the ordinate of B:
\[ c t_1 = \frac{c^2}{t_2} \]
Simplifying this gives:
\[ t_1 t_2 = c \]

3. **Equation of Tangents**:
The equation of the tangent to the hyperbola \(xy = c^2\) at point \(A\) is given by:
\[ \frac{x}{c t_1} + \frac{y}{\frac{c^2}{t_1}} = 2 \]
This simplifies to:
\[ \frac{x}{c t_1} + \frac{t_1 y}{c} = 2 \]
Multiplying through by \(c\) gives:
\[ x + t_1^2 y = 2c \]
Similarly, the equation of the tangent at point B is:
\[ x + t_2^2 y = 2c \]

4. **Find the Intersection of Tangents**:
To find the point of intersection of the tangents, we solve the two equations:
\[ x + t_1^2 y = 2c \quad (1) \]
\[ x + t_2^2 y = 2c \quad (2) \]
Subtracting (2) from (1) gives:
\[ (t_1^2 - t_2^2)y = 0 \]
This implies either \(y = 0\) or \(t_1^2 = t_2^2\).

5. **Case Analysis**:
If \(y = 0\), then substituting back into either tangent equation gives:
\[ x = 2c \]
If \(t_1^2 = t_2^2\), then \(t_1 = t_2\) or \(t_1 = -t_2\). However, since \(t_1 t_2 = c\), we can conclude that the points A and B are symmetric about the line \(y = x\).

6. **Locus of Intersection**:
The locus of the intersection point of the tangents will be along the line \(y = x\). Thus, we can express this as:
\[ y = x \]

### Conclusion
The locus of the point of intersection of the tangents at points A and B is given by the equation:
\[ \boxed{y = x} \]