AB, AC are tangents to a parabola `y^2=4ax; p_1, p_2, p_3` are the lengths of the perpendiculars from A, B, C on any tangents to the curve, then `p_2,p_1,p_3` are in:

Options

A. A.P.
B. G.P.
C. H.P.
D. none of these

Answer

To solve the problem, we need to analyze the given parabola and the points from which tangents are drawn. The parabola is given by the equation \( y^2 = 4ax \).

### Step-by-Step Solution:

1. **Identify Points on the Parabola:**
- Let points \( B \) and \( C \) on the parabola be represented as:
\[ B(t_1) = (at_1^2, 2at_1) \quad \text{and} \quad C(t_2) = (at_2^2, 2at_2) \]
- The point \( A \) from which the tangents are drawn can be represented as:
\[ A = (at_1t_2, a(t_1 + t_2)) \]

2. **Equation of Tangent to the Parabola:**
- The equation of the tangent to the parabola at point \( (at^2, 2at) \) can be expressed as:
\[ y = mx + \frac{a}{m} \]
- Rearranging gives:
\[ mx - y + \frac{a}{m} = 0 \]

3. **Calculate Perpendicular Distances:**
- The lengths of the perpendiculars from points \( A, B, C \) to the tangent line can be calculated using the formula for the distance from a point to a line:
\[ p = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
- For point \( B \):
\[ p_2 = \frac{|m(at_1^2) - (2at_1) + \frac{a}{m}|}{\sqrt{m^2 + 1}} \]
- Simplifying \( p_2 \):
\[ p_2 = \frac{a}{m} \cdot \frac{(mt_1 - 1)^2}{\sqrt{m^2 + 1}} \]

- For point \( C \):
\[ p_3 = \frac{|m(at_2^2) - (2at_2) + \frac{a}{m}|}{\sqrt{m^2 + 1}} \]
- Simplifying \( p_3 \):
\[ p_3 = \frac{a}{m} \cdot \frac{(mt_2 - 1)^2}{\sqrt{m^2 + 1}} \]

- For point \( A \):
\[ p_1 = \frac{|m(at_1t_2) - (a(t_1 + t_2)) + \frac{a}{m}|}{\sqrt{m^2 + 1}} \]
- Simplifying \( p_1 \):
\[ p_1 = \frac{a}{m} \cdot \frac{(mt_1t_2 - (t_1 + t_2) + 1)^2}{\sqrt{m^2 + 1}} \]

4. **Establish the Relationship:**
- From the expressions for \( p_1, p_2, p_3 \), we can derive:
\[ p_1^2 = p_2 \cdot p_3 \]
- This implies that \( p_1, p_2, p_3 \) are in **Geometric Progression (G.P.)**.

### Conclusion:
Thus, we conclude that \( p_1, p_2, p_3 \) are in G.P.