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  • 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 <img a

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 \mathrm{OAB} is \frac{98}{3} \sqrt{3}

then a^{2}-b^{2} is equal to:

 

 

Option: 1

\frac{392}{3}


Option: 2

\frac{196}{3}


Option: 3

98


Option: 4

196


Answers (1)

best_answer

In \: \Delta\: AOB

\begin{aligned} & \tan \frac{\pi}{6}=\frac{\mathrm{OB}}{\mathrm{OA}}=\frac{\mathrm{b}}{\mathrm{a}} \\ & \Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{b}}{\mathrm{a}} \\ & \Rightarrow a=\sqrt{3 b} \end{aligned}

\because area of triangle \Delta \mathrm{OAB}=\frac{1}{2} \times \mathrm{ab}=\frac{98}{3} \times \sqrt{3}

\begin{aligned} & \Rightarrow \frac{\sqrt{3} b^2}{2}=\frac{98}{\sqrt{3}} \\ & \Rightarrow b^2=\frac{98}{3} \times 2 \\ & \Rightarrow b=\sqrt{\frac{196}{3}} \\ & a=\sqrt{196} \\ & a^2-b^2=196-\frac{196}{3}=\frac{588-196}{3} \\ & \Rightarrow a^2-b^2=\frac{392}{3} \end{aligned}

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Gaurav

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