A straight line cuts off the intercepts OA = | vedclass

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Question

(A) The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$.The normal form of the line is $x \cos \alpha + y \sin \alpha = p$,w

Vedclass · Accepted Answer

$\frac{392}{3}$

Vedclass · Answer

(A) The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$.The normal form of the line is $x \cos \alpha + y \sin \alpha = p$,where $\alpha$ is the angle the perpendicular from the origin makes with the positive $x$-axis.Given the perpendicular makes an angle of $\frac{\pi}{6}$ with the positive $y$-axis,it makes an angle of $\alpha = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$ with the positive $x$-axis.Thus,the equation is $x \cos \frac{\pi}{3} + y \sin \frac{\pi}{3} = p$,which simplifies to $\frac{x}{2} + \frac{y \sqrt{3}}{2} = p$,or $\frac{x}{2p} + \frac{y}{2p/\sqrt{3}} = 1$.Comparing this with $\frac{x}{a} + \frac{y}{b} = 1$,we get $a = 2p$ and $b = \frac{2p}{\sqrt{3}}$.The area of $	riangle OAB$ is $\frac{1}{2} ab = \frac{98}{3} \sqrt{3}$.Substituting $a$ and $b$: $\frac{1}{2} (2p) \left( \frac{2p}{\sqrt{3}} ight) = \frac{98}{3} \sqrt{3} \implies \frac{2p^2}{\sqrt{3}} = \frac{98\sqrt{3}}{3} \implies 2p^2 = \frac{98 \cdot 3}{3} = 98 \implies p^2 = 49$.Now,$a^2 - b^2 = (2p)^2 - \left( \frac{2p}{\sqrt{3}} ight)^2 = 4p^2 - \frac{4p^2}{3} = \frac{8p^2}{3}$.Substituting $p^2 = 49$: $a^2 - b^2 = \frac{8 \cdot 49}{3} = \frac{392}{3}$.

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