A straight line cuts off the intercepts OA=a and OB=b on the...

Plotting the diagram of the given data we get,

Equation of straight line

Here given $OA = a \text{ and } OB = b$, so equation of line $AB$ in intercept form will be $\dfrac{x}{a} + \dfrac{y}{b} = 1 . . . . . . \left(1\right)$

And given perpendicular line of length $p$ is making $\dfrac{\pi}{3}$ angle with the $x -$axis,

So, the equation of line in normal form will be

$xcos\left(\dfrac{\pi}{3}\right) + ysin\left(\dfrac{\pi}{3}\right) = p$

$⇒ \dfrac{x}{2p} + \dfrac{y\sqrt{3}}{2p} = 1$

So, on comparing with the equation \left(1\right)$we get,</p><p>$a = 2p \text{ and } b = \dfrac{2p}{\sqrt{3}}$</p><p>Now area of the triangle$OAB$is </p><p>$\dfrac{1}{2} \times a \times b = \dfrac{98\sqrt{3}}{3}$</p><p>$⇒ \dfrac{1}{2} \times 2p \times \dfrac{2p}{\sqrt{3}} = \dfrac{98\sqrt{3}}{3}$</p><p>$⇒ {p}^{2} = 49$</p><p>Therefore,</p><p>${a}^{2} - {b}^{2} = 4{p}^{2} - \dfrac{4{p}^{2}}{3} = \dfrac{8}{3}{p}^{2}$</p><p>$⇒ {a}^{2} - {b}^{2} = \dfrac{8}{3}\left(49\right) = \dfrac{392}{3}