A straight line cuts off the intercepts OA = a and OB = b on the positive directions of x-axis and y-axis respectively. If the perpendicular from origin O to this line makes an angle of $$\frac{\pi}{6}$$ with positive direction of y-axis and the area of $$\triangle OAB$$ is $$\frac{98}{3}\sqrt{3}$$, then $$a^2 - b^2$$ is equal to:

To find the value of $$a^2 - b^2$$, we use the normal form of a straight line equation and relate its parameters to the given intercepts and area.

Let the normal form of the line equation be:

$$x \cos\alpha + y \sin\alpha = p$$

Here, $$p$$ is the perpendicular distance from the origin to the line, and $$\alpha$$ is the angle made by the perpendicular with the positive x-axis.

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Step 1: Determine the angle $$\alpha$$

We are given that the perpendicular from the origin to this line makes an angle of $$\frac{\pi}{6}$$ (or $$30^\circ$$) with the positive y-axis. Since the line cuts off positive intercepts on both axes, the perpendicular lies in the first quadrant.

Therefore, the angle made by the perpendicular with the positive x-axis is:

$$\alpha = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$$

Substituting $$\alpha = \frac{\pi}{3}$$ into the normal equation:

$$x \cos\left(\frac{\pi}{3}\right) + y \sin\left(\frac{\pi}{3}\right) = p$$

$$x\left(\frac{1}{2}\right) + y\left(\frac{\sqrt{3}}{2}\right) = p \implies \frac{x}{2p} + \frac{y}{\frac{2p}{\sqrt{3}}} = 1$$

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Step 2: Express the intercepts $$a$$ and $$b$$ in terms of $$p$$

Comparing this with the intercept form of a line equation $$\frac{x}{a} + \frac{y}{b} = 1$$, we get:

$$a = 2p$$

$$b = \frac{2p}{\sqrt{3}}$$

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Step 3: Solve for $$p$$ using the area of $$\triangle OAB$$

The area of the right-angled triangle formed by the intercepts is given by:

$$\text{Area} = \frac{1}{2} \times a \times b = \frac{98}{3}\sqrt{3}$$

Substituting our expressions for $$a$$ and $$b$$:

$$\frac{1}{2} \times (2p) \times \left(\frac{2p}{\sqrt{3}}\right) = \frac{98}{3}\sqrt{3}$$

$$\frac{2p^2}{\sqrt{3}} = \frac{98\sqrt{3}}{3}$$

Multiplying both sides by $$\sqrt{3}$$:

$$2p^2 = \frac{98 \times 3}{3} \implies 2p^2 = 98 \implies p^2 = 49$$

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Step 4: Calculate $$a^2 - b^2$$

Now, we write the expression for $$a^2 - b^2$$ in terms of $$p^2$$:

$$a^2 = (2p)^2 = 4p^2$$

$$b^2 = \left(\frac{2p}{\sqrt{3}}\right)^2 = \frac{4p^2}{3}$$

Evaluating the difference:

$$a^2 - b^2 = 4p^2 - \frac{4p^2}{3} = \frac{8p^2}{3}$$

Substituting $$p^2 = 49$$ into the equation:

$$a^2 - b^2 = \frac{8 \times 49}{3} = \frac{392}{3}$$

Therefore, the value of $$a^2 - b^2$$ is equal to $$\frac{392}{3}$$.