If a circle cuts a rectangular hyperbola xy=1 in four points P,Q,R,S and the parameters of these four points be `t_(1),t_(2),t_(3) and t_(4)` respectively and `-20t_(1)t_(2)t_(3)t_(4)=k`, then value of k equals

Options

A. 1
B. `-1`
C. `-4`
D. none

Answer

To find the value of \( k \) given that a circle intersects the rectangular hyperbola \( xy = 1 \) at four points \( P, Q, R, S \) with parameters \( t_1, t_2, t_3, t_4 \), we can follow these steps:

### Step 1: Understand the Parametric Form of the Hyperbola
The rectangular hyperbola \( xy = 1 \) can be represented in parametric form as:
\[ P(t) = (t, \frac{1}{t}) \]
Thus, the points of intersection can be expressed as:
\[ P(t_1) = (t_1, \frac{1}{t_1}), \quad P(t_2) = (t_2, \frac{1}{t_2}), \quad P(t_3) = (t_3, \frac{1}{t_3}), \quad P(t_4) = (t_4, \frac{1}{t_4}) \]

### Step 2: Circle Equation
The general equation of a circle can be expressed as:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
Substituting the parametric coordinates into the circle's equation gives:
\[ (t^2) + \left(\frac{1}{t^2}\right) + 2g(t) + 2f\left(\frac{1}{t}\right) + c = 0 \]
Multiplying through by \( t^2 \) to eliminate the fraction leads to:
\[ t^4 + 1 + 2gt^3 + 2ft + ct^2 = 0 \]

### Step 3: Form a Bi-Quadratic Equation
Rearranging the terms, we can write the equation as:
\[ t^4 + 2gt^3 + ct^2 + 2ft + 1 = 0 \]
This is a bi-quadratic equation in \( t \) with roots \( t_1, t_2, t_3, t_4 \).

### Step 4: Use Vieta's Formulas
From Vieta's formulas, we know that for a polynomial equation of the form:
\[ t^4 + at^3 + bt^2 + ct + d = 0 \]
the product of the roots \( t_1 t_2 t_3 t_4 \) is given by:
\[ t_1 t_2 t_3 t_4 = \frac{d}{a} \]
In our case, the equation is:
\[ t^4 + 2gt^3 + ct^2 + 2ft + 1 = 0 \]
Here, \( a = 2g \) and \( d = 1 \). Thus:
\[ t_1 t_2 t_3 t_4 = \frac{1}{1} = 1 \]

### Step 5: Calculate \( k \)
Given that:
\[ -20 t_1 t_2 t_3 t_4 = k \]
Substituting the value of \( t_1 t_2 t_3 t_4 \):
\[ k = -20 \cdot 1 = -20 \]

### Conclusion
Thus, the value of \( k \) is:
\[ \boxed{-20} \]