If the tangent and normal to a rectangular hyperbola cut off intercepts `x_(1)` and `x_(2)` on one axis and `y_(1)` and `y_(2)` on the other, then

Options

A. `x_1y_1+x_2y_2=0`
B. `x_1y_2+x_2y_1=0`
C. `x_1x_2+y_1y_2=0`
D. None of these

Answer

To solve the problem, we need to find the relationship between the intercepts \(x_1\), \(x_2\) on the x-axis and \(y_1\), \(y_2\) on the y-axis for the tangent and normal to a rectangular hyperbola.

### Step-by-Step Solution:

1. **Equation of the Rectangular Hyperbola**:
The equation of a rectangular hyperbola can be expressed as:
\[ xy = a^2 \]
or in another form:
\[ x^2 - y^2 = a^2 \]

2. **Parametric Representation**:
For a point \(P\) on the hyperbola, we can use the parametric equations:
\[ x = a \sec \theta, \quad y = a \tan \theta \]

3. **Equation of the Tangent**:
The equation of the tangent at point \(P\) is given by:
\[ x \sec \theta - y \tan \theta = a \]

4. **Finding Intercepts for Tangent**:
- To find the x-intercept \(x_1\), set \(y = 0\):
\[ x \sec \theta = a \implies x_1 = a \cos \theta \]
- To find the y-intercept \(y_1\), set \(x = 0\):
\[ -y \tan \theta = a \implies y_1 = -a \cot \theta \]

5. **Equation of the Normal**:
The equation of the normal at point \(P\) is given by:
\[ x \cos \theta + y \sin \theta = 2a \]

6. **Finding Intercepts for Normal**:
- To find the x-intercept \(x_2\), set \(y = 0\):
\[ x \cos \theta = 2a \implies x_2 = \frac{2a}{\cos \theta} = 2a \sec \theta \]
- To find the y-intercept \(y_2\), set \(x = 0\):
\[ y \sin \theta = 2a \implies y_2 = \frac{2a}{\sin \theta} = 2a \csc \theta \]

7. **Finding the Relationship**:
Now we will find the relationship between \(x_1, y_1, x_2, y_2\):
- The product of the intercepts gives:
\[ x_1 x_2 + y_1 y_2 = (a \cos \theta)(2a \sec \theta) + (-a \cot \theta)(2a \tan \theta) \]
- Simplifying this:
\[ = 2a^2 \cos \theta \sec \theta - 2a^2 \cot \theta \tan \theta \]
- Since \(\cos \theta \sec \theta = 1\) and \(\cot \theta \tan \theta = 1\):
\[ = 2a^2 - 2a^2 = 0 \]

8. **Final Result**:
Therefore, we have:
\[ x_1 x_2 + y_1 y_2 = 0 \]

### Conclusion:
The relationship between the intercepts is:
\[ x_1 x_2 + y_1 y_2 = 0 \]