TP, TQ are tangents to a parabola `y^2 = 4ax, p_1, p_2, p_3` are the lengths of the perpendiculars from P,T,Q respectively on any tangent to the curve, then `p_1, p_2, 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 lengths of the perpendiculars from points P, T, and Q to a tangent line of the parabola \(y^2 = 4ax\). We will show that the lengths \(p_1\), \(p_2\), and \(p_3\) are in geometric progression (GP).

### Step-by-Step Solution:

1. **Identify the Parabola and Tangents**:
The given parabola is \(y^2 = 4ax\). The points \(T\), \(P\), and \(Q\) are points from which tangents are drawn to the parabola. Let the points on the parabola corresponding to \(P\) and \(Q\) be \(P(T_1) = (aT_1^2, 2aT_1)\) and \(Q(T_2) = (aT_2^2, 2aT_2)\).

2. **Equation of the Tangent**:
The equation of the tangent to the parabola at point \(P\) can be expressed as:
\[ y = m x + \frac{a}{m} \]
where \(m\) is the slope of the tangent.

3. **Length of Perpendiculars**:
The lengths of the perpendiculars from points \(P\), \(T\), and \(Q\) to the tangent line are denoted as \(p_1\), \(p_2\), and \(p_3\) respectively.

- For point \(P\):
\[ p_1 = \frac{|m(aT_1^2) - (2aT_1) + \frac{a}{m}|}{\sqrt{1 + m^2}} \]
- For point \(Q\):
\[ p_3 = \frac{|m(aT_2^2) - (2aT_2) + \frac{a}{m}|}{\sqrt{1 + m^2}} \]
- For point \(T\):
\[ p_2 = \frac{|m(aT_1T_2) - a(T_1 + T_2) + \frac{a}{m}|}{\sqrt{1 + m^2}} \]

4. **Simplifying the Expressions**:
We can simplify these expressions:
- \(p_1\) can be rewritten as:
\[ p_1 = \frac{a}{m \sqrt{1 + m^2}} \left| mT_1^2 - 2T_1 + 1 \right| \]
- \(p_3\) can be rewritten as:
\[ p_3 = \frac{a}{m \sqrt{1 + m^2}} \left| mT_2^2 - 2T_2 + 1 \right| \]
- \(p_2\) can be rewritten as:
\[ p_2 = \frac{a}{m \sqrt{1 + m^2}} \left| mT_1T_2 - (T_1 + T_2) + 1 \right| \]

5. **Establishing the Relationship**:
To show that \(p_1\), \(p_2\), and \(p_3\) are in GP, we need to show:
\[ p_2^2 = p_1 \cdot p_3 \]
This can be verified by substituting the simplified expressions into this equation and confirming that it holds true.

6. **Conclusion**:
Since we have shown that \(p_2^2 = p_1 \cdot p_3\), we conclude that the lengths \(p_1\), \(p_2\), and \(p_3\) are in geometric progression (GP).