From a point A a varible straight line is drawn to cut the hyperbola `xy=c^2` at B and C. If a point P is chosen on BC such that AB,AP,AC are in G.P., then the locus of P is

Options

A. a straight line
B. parabola
C. a hyperbola with centre A
D. a hyperbola with centre O

Answer

To find the locus of point P such that AB, AP, and AC are in geometric progression (G.P.), we start with the hyperbola defined by the equation \(xy = c^2\) and a variable line cutting it at points B and C.

### Step-by-Step Solution:

1. **Define the Hyperbola and Points**:
The hyperbola is given by the equation \(xy = c^2\). Let point A be \((x_1, y_1)\). The line from point A intersects the hyperbola at points B and C.

2. **Equation of the Line**:
We assume the line can be represented in the parametric form:
\[ y - y_1 = m(x - x_1) \]
where \(m\) is the slope of the line.

3. **Substituting into the Hyperbola**:
Substitute \(y\) from the line equation into the hyperbola's equation:
\[ x(m(x - x_1) + y_1) = c^2 \]
Rearranging gives a quadratic equation in \(x\):
\[ mx^2 - mx_1x + (y_1x - c^2) = 0 \]

4. **Roots of the Quadratic**:
Let \(r_1\) and \(r_2\) be the roots of this quadratic equation, corresponding to the x-coordinates of points B and C. By Vieta's formulas, we have:
\[ r_1 + r_2 = \frac{mx_1 - y_1}{m} \]
\[ r_1r_2 = \frac{c^2}{m} \]

5. **Condition for G.P.**:
For points A, P, and C to be in G.P., the following condition must hold:
\[ AB \cdot AC = AP^2 \]
This can be expressed in terms of the distances:
\[ |r_1 - x_1| \cdot |r_2 - x_1| = |p - x_1|^2 \]

6. **Finding the Locus of P**:
By substituting the values of \(r_1\) and \(r_2\) into the G.P. condition and manipulating the equation, we can derive the locus of point P. The calculations yield:
\[ x_1y_1 - c^2 = hk - hy_1 - kx_1 + x_1y_1 \]
Rearranging gives:
\[ xy - x_1y - y_1x - c^2 = 0 \]

7. **Final Form of the Locus**:
The equation can be recognized as representing a hyperbola centered at point A \((x_1, y_1)\). Thus, the locus of point P is:
\[ xy - x_1y - y_1x - c^2 = 0 \]

### Conclusion:
The locus of point P is a hyperbola centered at point A.