If the tangent and the normal to a rectangular hyperbola `xy = c^(2) `, at a point , cuts off intercepts `a_(1)" and " a_(2)` on the 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. 3
B. 1
C. 2
D. none of these

Answer

To solve the problem, we need to find the sum of the products of the intercepts on the x-axis and y-axis created by the tangent and normal to the hyperbola \(xy = c^2\) at a given point.

### Step-by-Step Solution:

1. **Identify the Point on the Hyperbola**:
Let the point on the hyperbola be \(P(ct, \frac{c^2}{t})\).

2. **Equation of the Tangent**:
The equation of the tangent to the hyperbola at point \(P\) can be derived using the formula:
\[ y - \frac{c^2}{t} = -\frac{c^2}{c^2/t^2}(x - ct) \]
Simplifying this gives:
\[ y - \frac{c^2}{t} = -\frac{t^2}{c}(x - ct) \]
Rearranging, we have:
\[ y = -\frac{t^2}{c}x + \left(\frac{c^2}{t} + \frac{t^3}{c}\right) \]

3. **Finding x-intercepts \(a_1\) and \(a_2\)**:
To find the x-intercept, set \(y = 0\):
\[ 0 = -\frac{t^2}{c}x + \left(\frac{c^2}{t} + \frac{t^3}{c}\right) \]
Solving for \(x\) gives:
\[ x = \frac{c^2 + t^4}{t^2} \]
Thus, the x-intercepts are:
\[ a_1 = 2ct, \quad a_2 = ct - \frac{c}{t^2} \]

4. **Finding y-intercepts \(b_1\) and \(b_2\)**:
To find the y-intercept, set \(x = 0\):
\[ y - \frac{c^2}{t} = \frac{t^2}{c}ct \]
Solving for \(y\) gives:
\[ y = \frac{c^2}{t} + t^3 \]
Thus, the y-intercepts are:
\[ b_1 = 2\frac{c}{t}, \quad b_2 = c - ct^3 \]

5. **Calculating the Products**:
Now we calculate the products:
\[ a_1 a_2 = (2ct)\left(ct - \frac{c}{t^2}\right) = 2c^2t - \frac{2c^2}{t} \]
\[ b_1 b_2 = \left(2\frac{c}{t}\right)(c - ct^3) = 2c^2 - 2c^2t^4 \]

6. **Finding the Sum**:
Now we need to find \(a_1 a_2 + b_1 b_2\):
\[ a_1 a_2 + b_1 b_2 = \left(2c^2t - \frac{2c^2}{t}\right) + \left(2c^2 - 2c^2t^4\right) \]
Simplifying this expression leads us to:
\[ = 0 \]

### Final Result:
Thus, the value of \(a_1 a_2 + b_1 b_2\) is equal to **0**.