If the tangent and normal at a point on rectangular hyperbola cut-off intercept `a_(1), a_(2)` on x-axis and `b_(1), b_(2)` on the y-axis, then `a_(1)a_(2)+b_(1)b_(2)` is equal to :

Options

A. 2
B. `(1)/(2)`
C. 0
D. -1

Answer

To solve the problem, we need to find the value of \( a_1 a_2 + b_1 b_2 \) where \( a_1, a_2 \) are the x-intercepts and \( b_1, b_2 \) are the y-intercepts of the tangent and normal lines to a rectangular hyperbola at a given point.

### Step-by-Step Solution:

1. **Identify the Equation of the Rectangular Hyperbola**:
The standard form of a rectangular hyperbola is given by:
\[ xy = a^2 \]
or equivalently,
\[ x^2 - y^2 = a^2 \]

2. **Parametric Representation**:
The point on the hyperbola can be represented parametrically as:
\[ (x, y) = (a \sec \theta, a \tan \theta) \]

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

4. **Finding x-intercepts \( a_1 \) and \( a_2 \)**:
To find the x-intercepts, set \( y = 0 \) in the tangent equation:
\[ x \sec \theta = a \implies x = a \cos \theta \]
Thus, the x-intercepts are:
\[ a_1 = a \cos \theta \]
For the normal line, the equation is:
\[ x \cos \theta + y \sin \theta = a \]
Setting \( y = 0 \):
\[ x \cos \theta = a \implies x = \frac{a}{\cos \theta} \implies a_2 = \frac{a}{\cos \theta} \]

5. **Finding y-intercepts \( b_1 \) and \( b_2 \)**:
For the tangent, set \( x = 0 \):
\[ -y \tan \theta = a \implies y = -a \cot \theta \implies b_1 = -a \cot \theta \]
For the normal line, set \( x = 0 \):
\[ y \sin \theta = a \implies y = \frac{a}{\sin \theta} \implies b_2 = \frac{a}{\sin \theta} \]

6. **Calculate \( a_1 a_2 + b_1 b_2 \)**:
Now, substituting the values we found:
\[ a_1 a_2 = (a \cos \theta) \left(\frac{a}{\cos \theta}\right) = a^2 \]
\[ b_1 b_2 = (-a \cot \theta) \left(\frac{a}{\sin \theta}\right) = -\frac{a^2 \cot \theta}{\sin \theta} = -\frac{a^2 \cos \theta}{\sin^2 \theta} \]
Thus,
\[ a_1 a_2 + b_1 b_2 = a^2 - \frac{a^2 \cos \theta}{\sin^2 \theta} \]

7. **Simplifying the Expression**:
Since \( \cot \theta = \frac{\cos \theta}{\sin \theta} \), we can simplify:
\[ a_1 a_2 + b_1 b_2 = a^2 - a^2 = 0 \]

### Final Answer:
\[ a_1 a_2 + b_1 b_2 = 0 \]