Question

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 $\triangle O A B$ is $\frac{98}{3} \sqrt{3}$, then $a^2-b^2$ is equal to :

• A. $\frac{392}{3}$
• B. $\frac{196}{3}$
• C. 196
• D. 98

Answer 1

Answer: (A)

Equation of straight line : $\frac{ x }{ a }+\frac{ y }{ b }=1$
Or $x \cos \frac{\pi}{3}+y \sin \frac{\pi}{3}=p$
$\frac{x}{2}+\frac{y \sqrt{3}}{2}=p$
$\frac{x}{3 p}+\frac{y}{2 p}=1$
Comparing both : $a=2 p , b =\frac{2 p }{\sqrt{3}}$
Now area of $\triangle OAB =\frac{1}{2} \cdot ab =\frac{98}{3} \cdot \sqrt{3}$
$\frac{1}{2} \cdot 2 p \cdot \frac{2 p }{\sqrt{3}}=\frac{98}{3} \cdot \sqrt{3}$
$p ^2=49$
$a^2-b^2=4 p^2-\frac{4 p^2}{3}=\frac{2}{3} 4 p^2$
$=\frac{8}{3} \cdot 49=\frac{392}{3}$