If the normal at point 't' of the curve xy = `c^(2)`meets the curve again at point `'t'_(1)`, then prove that
`t^(3)* t_(1) =- 1`.

Answer

To prove that \( t^3 \cdot t_1 = -1 \) for the given curve \( xy = c^2 \), we will follow these steps:

### Step 1: Identify the points on the curve
The curve is given by the equation \( xy = c^2 \). For a point on the curve corresponding to the parameter \( t \), we can express the coordinates as:
\[ (x, y) = (ct, \frac{c}{t}) \]

### Step 2: Find the slope of the tangent
To find the slope of the normal at the point \( (ct, \frac{c}{t}) \), we first need to find the derivative \( \frac{dy}{dx} \). We differentiate the equation \( xy = c^2 \) implicitly:
\[ \frac{d}{dx}(xy) = \frac{d}{dx}(c^2) \]
Using the product rule, we have:
\[ x \frac{dy}{dx} + y = 0 \implies \frac{dy}{dx} = -\frac{y}{x} \]
Substituting \( y = \frac{c}{t} \) and \( x = ct \):
\[ \frac{dy}{dx} = -\frac{\frac{c}{t}}{ct} = -\frac{1}{t^2} \]

### Step 3: Determine the slope of the normal
The slope of the normal is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = -\frac{1}{\left(-\frac{1}{t^2}\right)} = t^2 \]

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

### Step 5: Find the intersection with the curve again
We need to find where this normal intersects the curve \( xy = c^2 \) again. Substituting \( y \) from the normal's equation into the curve's equation:
\[ x \left(t^2 x - ct^3 + \frac{c}{t}\right) = c^2 \]
Expanding and rearranging gives:
\[ t^2 x^2 - ct^3 x + \frac{c}{t} x - c^2 = 0 \]
This is a quadratic equation in \( x \).

### Step 6: Use the quadratic formula
Let \( A = t^2 \), \( B = -ct^3 + \frac{c}{t} \), and \( C = -c^2 \). The roots of the quadratic equation can be found using the quadratic formula:
\[ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]
Since we already have one root at \( x = ct \), we can denote the other root as \( x = ct_1 \).

### Step 7: Relate the roots
By Vieta's formulas, the product of the roots \( ct \cdot ct_1 = \frac{C}{A} \):
\[ ct \cdot ct_1 = \frac{-c^2}{t^2} \]
This simplifies to:
\[ c^2 tt_1 = -c^2 \implies tt_1 = -1 \]
Thus, multiplying both sides by \( t^2 \):
\[ t^3 t_1 = -1 \]

### Conclusion
We have shown that:
\[ t^3 t_1 = -1 \]
This completes the proof.

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