Let $$a$$ and $$b$$ be any two numbers satisfying $$\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{4}$$. Then, the foot of perpendicular from the origin on the variable line $$\frac{x}{a} + \frac{y}{b} = 1$$ lies on:

We are given that $$a$$ and $$b$$ satisfy the equation $$\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{4}$$. The variable line is given by $$\frac{x}{a} + \frac{y}{b} = 1$$. We need to find the locus of the foot of the perpendicular from the origin $$(0, 0)$$ to this line.

First, rewrite the line equation in standard form. Multiply both sides by $$ab$$ to clear denominators:

$$b x + a y = ab$$

So, the line is $$b x + a y - ab = 0$$. Here, $$A = b$$, $$B = a$$, and $$C = -ab$$.

The foot of the perpendicular from $$(0, 0)$$ to the line $$Ax + By + C = 0$$ can be found using the formulas:

$$h = \frac{-A C}{A^2 + B^2}, \quad k = \frac{-B C}{A^2 + B^2}$$

Substituting $$A = b$$, $$B = a$$, and $$C = -ab$$:

$$h = \frac{-b \cdot (-ab)}{b^2 + a^2} = \frac{a b^2}{a^2 + b^2}$$

$$k = \frac{-a \cdot (-ab)}{a^2 + b^2} = \frac{a^2 b}{a^2 + b^2}$$

Now, compute $$h^2 + k^2$$:

$$h^2 + k^2 = \left( \frac{a b^2}{a^2 + b^2} \right)^2 + \left( \frac{a^2 b}{a^2 + b^2} \right)^2 = \frac{a^2 b^4}{(a^2 + b^2)^2} + \frac{a^4 b^2}{(a^2 + b^2)^2} = \frac{a^2 b^4 + a^4 b^2}{(a^2 + b^2)^2}$$

Factor the numerator:

$$a^2 b^4 + a^4 b^2 = a^2 b^2 (b^2 + a^2) = a^2 b^2 (a^2 + b^2)$$

So,

$$h^2 + k^2 = \frac{a^2 b^2 (a^2 + b^2)}{(a^2 + b^2)^2} = \frac{a^2 b^2}{a^2 + b^2}$$

We are given the condition $$\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{4}$$. Rewrite this as:

$$\frac{b^2 + a^2}{a^2 b^2} = \frac{1}{4}$$

Multiply both sides by $$a^2 b^2$$:

$$a^2 + b^2 = \frac{a^2 b^2}{4}$$

Rearrange to solve for $$\frac{a^2 b^2}{a^2 + b^2}$$:

$$\frac{a^2 b^2}{a^2 + b^2} = 4$$

Substitute this into the expression for $$h^2 + k^2$$:

$$h^2 + k^2 = 4$$

This equation represents a circle centered at the origin $$(0, 0)$$ with radius 2.

Comparing with the options:

A. A circle of radius = 2

B. A hyperbola with each semi-axis = $$\sqrt{2}$$

C. A hyperbola with each semi-axis = 2

D. A circle of radius = $$\sqrt{2}$$

Hence, the correct answer is Option A.