A straight line cuts off the intercepts OA=a and OB=b on the positive directions of x-axis…

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 $\dfrac{\pi}6$ with positive direction of $y$-axis and the area of $	riangle OAB$ is $\dfrac{98}3\sqrt3$, then $a^2-b^2$ is equal to:

Accepted Answer

$${1 \over 2}ab = {{98\sqrt 3 } \over 3}$$

$$ \Rightarrow \sqrt 3 ab = 196$$ ..... (i)

$$OP = OB\cos 30^\circ  = OA\cos 60^\circ $$

$$ \Rightarrow {{b\sqrt 3 } \over 2} = {a \over 2}$$

$$ \Rightarrow \sqrt 3 b = a$$ ..... (ii)

By (i) and (ii)

${a^2} = 196$

$a = 14$

$${b^2} = {{{a^2}} \over 3}$$

$${a^2} - {b^2} = {{2{a^2}} \over 3} = {{392} \over 3}$$

Solutions · 1