The normal at anypoint P (ct,c/t) on the curve xy = `c^2` meets the curve at `Q (ct_1 ,c // t_1 )`, then `t_1 =`

Options

A. `-t`
B. `1//t^2`
C. `-1//t^3`
D. none

Answer

To solve the problem, we need to find the value of \( t_1 \) where the normal at point \( P(ct, \frac{c}{t}) \) on the curve \( xy = c^2 \) meets the curve again at point \( Q(ct_1, \frac{c}{t_1}) \).

### Step-by-Step Solution:

1. **Identify the Curve and Point P**:
The given curve is \( xy = c^2 \). The point \( P \) is given as \( P(ct, \frac{c}{t}) \).

2. **Differentiate the Curve**:
To find the slope of the tangent at point \( P \), we differentiate the equation \( xy = c^2 \) with respect to \( x \):
\[ y + x \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{y}{x} \]

3. **Calculate the Slope at Point P**:
Substitute the coordinates of point \( P \) into the derivative:
\[ \frac{dy}{dx} \bigg|_P = -\frac{\frac{c}{t}}{ct} = -\frac{1}{t^2} \]
Thus, the slope of the tangent at \( P \) is \( -\frac{1}{t^2} \).

4. **Find the Slope of the Normal**:
The slope of the normal \( m_1 \) is the negative reciprocal of the slope of the tangent:
\[ m_1 = -\frac{1}{\left(-\frac{1}{t^2}\right)} = t^2 \]

5. **Write the Equation of the Normal**:
The equation of the normal line at point \( P \) can be written as:
\[ y - \frac{c}{t} = t^2 \left(x - ct\right) \]
Rearranging gives:
\[ y = t^2 x - ct^3 + \frac{c}{t} \]

6. **Substitute Point Q into the Normal Equation**:
Since point \( Q \) is \( (ct_1, \frac{c}{t_1}) \), substitute these coordinates into the normal equation:
\[ \frac{c}{t_1} = t^2 (ct_1) - ct^3 + \frac{c}{t} \]

7. **Simplify the Equation**:
Multiply through by \( t_1 \) to eliminate the fraction:
\[ c = t^2 ct_1^2 - ct^3 t_1 + \frac{c}{t} t_1 \]
Rearranging gives:
\[ 0 = t^2 ct_1^2 - ct^3 t_1 + \frac{c}{t} t_1 - c \]

8. **Factor Out c**:
Factor out \( c \) (assuming \( c \neq 0 \)):
\[ 0 = t^2 t_1^2 - t^3 t_1 + \frac{1}{t} t_1 - 1 \]

9. **Rearranging**:
Rearranging gives:
\[ t^2 t_1^2 - (t^3 - \frac{1}{t}) t_1 - 1 = 0 \]

10. **Solve for \( t_1 \)**:
This is a quadratic equation in \( t_1 \). Using the quadratic formula:
\[ t_1 = \frac{(t^3 - \frac{1}{t}) \pm \sqrt{(t^3 - \frac{1}{t})^2 + 4t^2}}{2t^2} \]

11. **Final Result**:
After simplification, we find:
\[ t_1 = \frac{1}{t^3} \]

### Final Answer:
\[ t_1 = \frac{1}{t^3} \]