If 2p is the perpendicular distance from the origin to the line `x/a+y/b=1` then `a^(2),8p^(2),b^(2)` are in

Options

A. A.P
B. G.P
C. H.P
D. None

Answer

To solve the problem, we need to find the relationship between \( a^2 \), \( 8p^2 \), and \( b^2 \) given that \( 2p \) is the perpendicular distance from the origin to the line \( \frac{x}{a} + \frac{y}{b} = 1 \).

### Step-by-Step Solution:

1. **Identify the line equation**: The given line is \( \frac{x}{a} + \frac{y}{b} = 1 \). We can rewrite this in the standard form:
\[ \frac{x}{a} + \frac{y}{b} - 1 = 0 \]

2. **Calculate the perpendicular distance from the origin**: The formula for the perpendicular distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
Here, \( A = \frac{1}{a} \), \( B = \frac{1}{b} \), and \( C = -1 \). The coordinates of the origin are \( (0, 0) \).

Substituting these values into the formula gives:
\[ d = \frac{\left| \frac{0}{a} + \frac{0}{b} - 1 \right|}{\sqrt{\left(\frac{1}{a}\right)^2 + \left(\frac{1}{b}\right)^2}} = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} \]

3. **Set the distance equal to \( 2p \)**:
\[ 2p = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} \]

4. **Square both sides**:
\[ (2p)^2 = \frac{1}{\frac{1}{a^2} + \frac{1}{b^2}} \]
This simplifies to:
\[ 4p^2 = \frac{1}{\frac{1}{a^2} + \frac{1}{b^2}} \]

5. **Reciprocate the equation**:
\[ \frac{1}{4p^2} = \frac{1}{a^2} + \frac{1}{b^2} \]

6. **Find the harmonic mean (HM)**: The harmonic mean of \( a^2 \) and \( b^2 \) is given by:
\[ HM = \frac{2ab}{a + b} \]
However, we can express it in terms of reciprocals:
\[ HM = \frac{2}{\frac{1}{a^2} + \frac{1}{b^2}} = \frac{2}{\frac{1}{4p^2}} = 8p^2 \]

7. **Conclusion**: Since \( 8p^2 \) is the harmonic mean of \( a^2 \) and \( b^2 \), we conclude that \( a^2, 8p^2, b^2 \) are in harmonic progression (HP).

### Final Answer:
Thus, \( a^2, 8p^2, b^2 \) are in harmonic progression (HP).