Question

A straight line cuts of the intercept OA =a and OB=b on the positive direction of x-axis and y axis respectively. If the perpendicular from origin o to this like makes an angle of π/6 with positive direction of y axis and the area of triangle OAB is 98+3√3/3 then a²-b² is equal to

Answer 1

In \( \Delta A O B\)

\(\tan \frac{\pi}{6}=\frac{\mathrm{OB}}{\mathrm{OA}}=\frac{\mathrm{b}}{\mathrm{a}} \)

\(\Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{b}}{\mathrm{a}} \)

\(\Rightarrow a=\sqrt{3 b}\)

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

\(\Rightarrow \frac{\sqrt{3} b^2}{2}=\frac{98}{\sqrt{3}}\)

\(\Rightarrow b^2=\frac{98}{3} \times 2 \)

\(\Rightarrow b=\sqrt{\frac{196}{3}}\)

\(a=\sqrt{196}\)

\(a^2-b^2=196-\frac{196}{3}=\frac{588-196}{3}\)

\(\Rightarrow a^2-b^2=\frac{392}{3}\)