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

The Correct Option is A

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