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Q.

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 form origin O to this line makes an angle of $\frac{π}{6}$ with positive direction of y-axis and the area of $Δ O A B$ is $\frac{98 \sqrt{3}}{3}$ then $a^{2} - b^{2} =$

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a

$\frac{196}{3}$

b

$196$

c

$98$

d

$\frac{392}{3}$

answer is D.

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Detailed Solution

$\tan \frac{π}{6} = \frac{b}{a} \Rightarrow a = b \sqrt{3}$
$\begin{array}{l} A = \frac{98 \sqrt{3}}{3} \\ \frac{1}{2} a b = \frac{98 \sqrt{3}}{3} \\ b . \sqrt{3} . b = \frac{196 \sqrt{3}}{3} \Rightarrow b^{2} = \frac{196}{3} \\ a^{2} - b^{2} = 3 b^{2} - b^{2} = 2 b^{2} = \frac{392}{3} \end{array}$
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