If the normal at `(ct_(1),c//t_(1))` on the curve `xy=c^(2)` meets the curve again at the point `(ct_(2),c//t_(2))` then

Options

A. `t_(2)=-(1)/(t_(1)^(3))`
B. `t_(2)=-(1)/(t_(1))`
C. `t_(2)=(1)/(t_(1)^(2))`
D. none

Answer

To solve the problem, we need to analyze the normal line at the point \((ct_1, \frac{c}{t_1})\) on the hyperbola given by the equation \(xy = c^2\) and find where this normal intersects the hyperbola again at the point \((ct_2, \frac{c}{t_2})\).

### Step-by-Step Solution:

1. **Identify the point on the hyperbola**:
The point on the hyperbola is given as \((ct_1, \frac{c}{t_1})\).

2. **Find the slope of the tangent line**:
The equation of the hyperbola is \(xy = c^2\). To find the slope of the tangent at the point \((ct_1, \frac{c}{t_1})\), we can differentiate the equation implicitly:
\[ y + x \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{y}{x} \]
At the point \((ct_1, \frac{c}{t_1})\):
\[ \frac{dy}{dx} = -\frac{\frac{c}{t_1}}{ct_1} = -\frac{1}{t_1^2} \]

3. **Find the slope of the normal line**:
The slope of the normal line is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = t_1^2 \]

4. **Write the equation of the normal line**:
Using the point-slope form of the equation of a line, the equation of the normal line at the point \((ct_1, \frac{c}{t_1})\) is:
\[ y - \frac{c}{t_1} = t_1^2 \left(x - ct_1\right) \]
Rearranging gives:
\[ y = t_1^2 x - ct_1^3 + \frac{c}{t_1} \]

5. **Substitute the normal line into the hyperbola equation**:
We need to find where this normal line intersects the hyperbola again. Substitute \(y\) from the normal line into the hyperbola equation \(xy = c^2\):
\[ x(t_1^2 x - ct_1^3 + \frac{c}{t_1}) = c^2 \]
Simplifying gives:
\[ t_1^2 x^2 - ct_1^3 x + \frac{c}{t_1} x - c^2 = 0 \]

6. **Rearranging the quadratic equation**:
This is a quadratic equation in \(x\):
\[ t_1^2 x^2 + \left(\frac{c}{t_1} - ct_1^3\right)x - c^2 = 0 \]

7. **Using Vieta's formulas**:
Let the roots of this quadratic be \(x_1 = ct_1\) and \(x_2 = ct_2\). By Vieta's formulas:
\[ x_1 + x_2 = -\frac{b}{a} \quad \text{and} \quad x_1 x_2 = \frac{c}{a} \]
From the first equation:
\[ ct_1 + ct_2 = -\frac{\frac{c}{t_1} - ct_1^3}{t_1^2} \]
From the second equation:
\[ ct_1 \cdot ct_2 = -\frac{c^2}{t_1^2} \]

8. **Finding \(t_2\)**:
We can express \(t_2\) in terms of \(t_1\):
\[ t_1^3 t_2 + 1 = 0 \implies t_2 = -\frac{1}{t_1^3} \]

### Final Result:
Thus, we find that:
\[ t_2 = -\frac{1}{t_1^3} \]