If p is the length of perpendicular from the origin on the line `x/a+y/b=1` and `a^(2),p^(2),b^(2)` are in A.P., then

(A) `a^(4)+b^(4)=0`
(B) `a^(4)-b^(4)=0`
(C) `a^(2)+b^(2)=0`
(D) `a^(2)-b^(2)=0`

Options

A. `a^(4)+b^(4)=0`
B. `a^(4)-b^(4)=0`
C. `a^(2)+b^(2)=0`
D. `a^(2)-b^(2)=0`

Answer

To solve the problem, we need to find the relationship between \( a^2 \), \( p^2 \), and \( b^2 \) when they are in Arithmetic Progression (A.P.).

### Step-by-Step Solution:

1. **Understanding the Line Equation**:
The given line is \( \frac{x}{a} + \frac{y}{b} = 1 \). We can rearrange this to the standard form:
\[ bx + ay = ab \]

2. **Finding the Length of Perpendicular from the Origin**:
The formula for the distance \( p \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ p = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our line \( bx + ay - ab = 0 \), we have \( A = b \), \( B = a \), and \( C = -ab \). Substituting \( (x_0, y_0) = (0, 0) \):
\[ p = \frac{|b(0) + a(0) - ab|}{\sqrt{b^2 + a^2}} = \frac{ab}{\sqrt{a^2 + b^2}} \]

3. **Finding \( p^2 \)**:
Squaring both sides, we get:
\[ p^2 = \left(\frac{ab}{\sqrt{a^2 + b^2}}\right)^2 = \frac{a^2b^2}{a^2 + b^2} \]

4. **Setting Up the A.P. Condition**:
Since \( a^2, p^2, b^2 \) are in A.P., we can use the property of A.P.:
\[ 2p^2 = a^2 + b^2 \]
Substituting the expression for \( p^2 \):
\[ 2 \left(\frac{a^2b^2}{a^2 + b^2}\right) = a^2 + b^2 \]

5. **Cross Multiplying**:
Cross-multiplying gives:
\[ 2a^2b^2 = (a^2 + b^2)(a^2 + b^2) \]
Expanding the right-hand side:
\[ 2a^2b^2 = a^4 + 2a^2b^2 + b^4 \]

6. **Rearranging the Equation**:
Rearranging the equation:
\[ 0 = a^4 + b^4 + 2a^2b^2 - 2a^2b^2 \]
This simplifies to:
\[ a^4 + b^4 = 0 \]

7. **Conclusion**:
Since \( a^4 + b^4 = 0 \), the only way this holds true is if both \( a \) and \( b \) are zero. Thus, the correct option is:
\[ \text{(A) } a^4 + b^4 = 0 \]