If the normal at the point `P(x_1y_1),i=1.2,3,4` on the hyperbola `xy=c^2` are concurrent at the point Q(h, k), then `((x_1+x_2+x_3+x_4)(y_1+y_2+y_3+y_4))/(x_1x_2x_3x_4)` is:

Options

A. `(hk)/(c^2)`
B. `(hk)/(c^4)`
C. `-(hk)/(c^2)`
D. `-(hk)/(c^4)`

Answer

To solve the problem, we need to find the value of \(\frac{(x_1 + x_2 + x_3 + x_4)(y_1 + y_2 + y_3 + y_4)}{x_1 x_2 x_3 x_4}\) given that the normals at points \(P(x_1, y_1), P(x_2, y_2), P(x_3, y_3), P(x_4, y_4)\) on the hyperbola \(xy = c^2\) are concurrent at point \(Q(h, k)\).

### Step 1: Identify the points on the hyperbola
The hyperbola is given by the equation \(xy = c^2\). We can express the points \(P\) on the hyperbola in terms of a parameter \(t\):
\[ P(t) = (ct, \frac{c^2}{t}) \]
Thus, we can denote the points as:
- \(P_1 = (ct_1, \frac{c^2}{t_1})\)
- \(P_2 = (ct_2, \frac{c^2}{t_2})\)
- \(P_3 = (ct_3, \frac{c^2}{t_3})\)
- \(P_4 = (ct_4, \frac{c^2}{t_4})\)

### Step 2: Find the equation of the normal at point \(P(t)\)
The slope of the tangent to the hyperbola at point \(P(t)\) is given by:
\[ \text{slope} = -\frac{y}{x} = -\frac{c^2/t}{ct} = -\frac{c}{t} \]
Thus, the slope of the normal is:
\[ \text{slope of normal} = \frac{t}{c} \]
The equation of the normal at point \(P(t)\) can be written as:
\[ y - \frac{c^2}{t} = \frac{t}{c}(x - ct) \]
Rearranging gives:
\[ y = \frac{t}{c}x - \frac{t^2}{c} + \frac{c^2}{t} \]
This can be rewritten as:
\[ ty - tx + c^2 - t^2 = 0 \]

### Step 3: Substitute the point \(Q(h, k)\)
Since the normals are concurrent at point \(Q(h, k)\), substituting \(x = h\) and \(y = k\) into the normal equation gives:
\[ th - tk + c^2 - t^2 = 0 \]
This leads to the equation:
\[ th - tk + c^2 = t^2 \]
Rearranging gives:
\[ t^2 - th + tk - c^2 = 0 \]

### Step 4: Analyze the quadratic equation
The equation \(t^2 - th + tk - c^2 = 0\) is a quadratic in \(t\) with roots \(t_1, t_2, t_3, t_4\). By Vieta's formulas, we know:
- The sum of the roots \(t_1 + t_2 + t_3 + t_4 = h\)
- The product of the roots \(t_1 t_2 t_3 t_4 = -c^2\)

### Step 5: Calculate \(x_1 + x_2 + x_3 + x_4\) and \(y_1 + y_2 + y_3 + y_4\)
Using the expressions for \(x\) and \(y\):
\[ x_1 + x_2 + x_3 + x_4 = c(t_1 + t_2 + t_3 + t_4) = ch \]
\[ y_1 + y_2 + y_3 + y_4 = \frac{c^2}{t_1} + \frac{c^2}{t_2} + \frac{c^2}{t_3} + \frac{c^2}{t_4} = c^2\left(\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} + \frac{1}{t_4}\right) \]
Using the relationship from Vieta's formulas, we have:
\[ \frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3} + \frac{1}{t_4} = \frac{t_2 t_3 t_4 + t_1 t_3 t_4 + t_1 t_2 t_4 + t_1 t_2 t_3}{t_1 t_2 t_3 t_4} = \frac{k}{-c^2} \]
Thus:
\[ y_1 + y_2 + y_3 + y_4 = c^2 \cdot \frac{k}{-c^2} = -k \]

### Step 6: Calculate the final expression
Now substituting these into the expression we want to evaluate:
\[ \frac{(x_1 + x_2 + x_3 + x_4)(y_1 + y_2 + y_3 + y_4)}{x_1 x_2 x_3 x_4} = \frac{(ch)(-k)}{-c^4} = \frac{hk}{c^4} \]

### Final Result
Thus, the value is:
\[ \frac{hk}{c^4} \]