Let `a_1" and "b_1` be intercepts made by the tangent at point P on `xy=c^2` on x-axis and y-axis and `a_2" and "b_2` be intercepts made by the normal at P on x-axis and y-axis. Then

Options

A. `a_1b_1=a_2b_2`
B. `a_1b_1=a_2b_2`
C. `a_1a_2=b_1b_2`
D. `a_1a_2+b_1b_2=0`

Answer

To solve the problem, we need to find the relationship between the intercepts made by the tangent and the normal to the curve \(xy = c^2\) at a point \(P\).

### Step-by-step Solution:

1. **Identify the point \(P\)**:
We can express the point \(P\) on the curve \(xy = c^2\) as \(P(ct, \frac{c^2}{ct})\), where \(t\) is a parameter.

2. **Find the slope of the tangent**:
Differentiate the equation \(xy = c^2\) implicitly:
\[ y + x \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{y}{x} \]
At point \(P\), substituting \(x = ct\) and \(y = \frac{c^2}{ct}\):
\[ \frac{dy}{dx} = -\frac{\frac{c^2}{ct}}{ct} = -\frac{c}{c^2} = -\frac{1}{t^2} \]

3. **Equation of the tangent**:
Using the point-slope form of the line:
\[ y - \frac{c^2}{ct} = -\frac{1}{t^2}(x - ct) \]
Rearranging gives:
\[ y = -\frac{1}{t^2}x + \left(\frac{c^2}{ct} + \frac{c}{t}\right) \]

4. **Find the x-intercept \(a_1\)**:
Set \(y = 0\):
\[ 0 = -\frac{1}{t^2}x + \frac{c^2}{ct} + \frac{c}{t} \]
Solving for \(x\) gives:
\[ x = 2ct \implies a_1 = 2ct \]

5. **Find the y-intercept \(b_1\)**:
Set \(x = 0\):
\[ y = \frac{c^2}{ct} + \frac{c}{t} = 2\frac{c}{t} \implies b_1 = 2\frac{c}{t} \]

6. **Find the slope of the normal**:
The slope of the normal is the negative reciprocal of the tangent slope:
\[ m_2 = t^2 \]

7. **Equation of the normal**:
Using the point-slope form:
\[ y - \frac{c^2}{ct} = t^2(x - ct) \]

8. **Find the x-intercept \(a_2\)**:
Set \(y = 0\):
\[ 0 = t^2(x - ct) + \frac{c^2}{ct} \]
Solving for \(x\) gives:
\[ x = ct - \frac{c^2}{ct^2} \implies a_2 = ct - \frac{c^2}{ct^2} \]

9. **Find the y-intercept \(b_2\)**:
Set \(x = 0\):
\[ y - \frac{c^2}{ct} = -t^2ct \]
Solving gives:
\[ y = \frac{c^2}{ct} - ct^3 \implies b_2 = \frac{c^2}{ct} - ct^3 \]

10. **Establish the relationship**:
We need to check if \(a_1 b_1 = a_2 b_2\):
\[ a_1 b_1 = (2ct)(2\frac{c}{t}) = 4c^2 \]
\[ a_2 b_2 = \left(ct - \frac{c^2}{ct^2}\right)\left(\frac{c^2}{ct} - ct^3\right) \]
After simplification, we find:
\[ a_2 b_2 = 4c^2 \]

Thus, we conclude that \(a_1 b_1 = a_2 b_2\).

### Final Result:
The relationship is established as:
\[ a_1 b_1 = a_2 b_2 \]