A tangent to `x^(2)=32y` meets `xy=c^(2)` at P & Q. The locus of mid-point of PQ is a parabola whose latus rectum is-

Options

A. 3
B. 4
C. 5
D. 6

Explanation

To solve the problem step by step, we need to find the locus of the midpoint of points P and Q where a tangent to the parabola \(x^2 = 32y\) intersects the hyperbola \(xy = c^2\).

### Step 1: Identify the equations
The given equations are:
1. Parabola: \(x^2 = 32y\)
2. Hyperbola: \(xy = c^2\)

### Step 2: Write the equation of the tangent to the parabola
The standard form of the parabola \(x^2 = 4Ay\) gives us \(4A = 32\), hence \(A = 8\). The equation of the tangent to the parabola at point \((x_1, y_1)\) is given by:
\[ y = mx - 8m^2 \]
where \(m\) is the slope of the tangent.

### Step 3: Find the coordinates of the points of intersection (P and Q)
The tangent line intersects the hyperbola \(xy = c^2\). Substituting \(y\) from the tangent equation into the hyperbola gives:
\[ x(mx - 8m^2) = c^2 \]
This simplifies to:
\[ mx^2 - 8m^2x - c^2 = 0 \]
This is a quadratic equation in \(x\).

### Step 4: Use the midpoint formula
Let the roots of the quadratic equation be \(x_1\) and \(x_2\) (the x-coordinates of points P and Q). The midpoint \(M\) of \(PQ\) is given by:
\[ M_x = \frac{x_1 + x_2}{2}, \quad M_y = \frac{y_1 + y_2}{2} \]
Using Vieta's formulas, we know:
\[ x_1 + x_2 = -\frac{-8m^2}{m} = \frac{8m^2}{m} = 8m \]
Thus,
\[ M_x = \frac{8m}{2} = 4m \]

### Step 5: Find \(M_y\) in terms of \(M_x\)
Substituting \(M_x\) back into the tangent equation to find \(M_y\):
\[ M_y = m(4m) - 8m^2 = 4m^2 - 8m^2 = -4m^2 \]

### Step 6: Eliminate \(m\) to find the locus
Now we have:
\[ M_x = 4m \implies m = \frac{M_x}{4} \]
Substituting this into the equation for \(M_y\):
\[ M_y = -4\left(\frac{M_x}{4}\right)^2 = -\frac{M_x^2}{4} \]
This represents the equation of a parabola:
\[ M_y = -\frac{1}{4}M_x^2 \]

### Step 7: Identify the latus rectum
The standard form of a parabola is \(y^2 = 4ax\). Here, we have:
\[ M_y = -\frac{1}{4}M_x^2 \implies M_y = -\frac{1}{4}x^2 \]
The latus rectum \(L\) of a parabola is given by \(4a\). In this case, since the coefficient of \(M_x^2\) is \(-\frac{1}{4}\), we can see that:
\[ L = 4 \times \frac{1}{4} = 1 \]

### Final Answer
Thus, the latus rectum of the parabola whose locus is the midpoint of PQ is:
\[ \boxed{1} \]