JEE Mathematics 2023 Question with Solution

Given: The line cuts intercepts $a$ and $b$ on the positive $x$- and $y$-axes. The perpendicular from the origin makes an angle of $\frac{\pi}{6}$ with the positive direction of the $y$-axis. The area of $\triangle OAB$ is $\frac{98}{\sqrt{3}}$ as used in the solution working.

Find: $a^2-b^2$.

The intercept form of the line is

$$\frac{x}{a}+\frac{y}{b}=1$$

The perpendicular makes angle $\frac{\pi}{3}$ with the positive $x$-axis, so the perpendicular form is

$$x\cos\frac{\pi}{3}+y\sin\frac{\pi}{3}=p$$

Substituting the trigonometric values,

$$\frac{x}{2}+\frac{\sqrt{3}y}{2}=p$$

Rearranging into intercept form,

$$\frac{x}{2p}+\frac{y}{2p/\sqrt{3}}=1$$

Comparing with $\frac{x}{a}+\frac{y}{b}=1$, we get

$$a=2p, \qquad b=\frac{2p}{\sqrt{3}}$$

Now area of $\triangle OAB$ is

$$\frac{1}{2}ab$$

So,

$$\frac{1}{2}(2p)\left(\frac{2p}{\sqrt{3}}\right)=\frac{98}{\sqrt{3}}$$

This gives

$$\frac{2p^2}{\sqrt{3}}=\frac{98}{\sqrt{3}}$$

Hence,

$$p^2=49$$

Now,

$$a^2-b^2=(2p)^2-\left(\frac{2p}{\sqrt{3}}\right)^2$$ $$=4p^2-\frac{4p^2}{3}$$ $$=\frac{8p^2}{3}$$

Substituting $p^2=49$,

$$a^2-b^2=\frac{8}{3}\cdot 49=\frac{392}{3}$$

Therefore, the correct option is A.

The solution declares Option A and concludes $\frac{392}{3}$. However, the intermediate text in the solution contains inconsistent lines such as $a=2p$ and $b=2p\sqrt{3}$, and one rearrangement is also written incorrectly. Using the angle information correctly, the perpendicular makes angle $\frac{\pi}{3}$ with the positive $x$-axis, so the valid comparison gives

$$a=2p, \qquad b=\frac{2p}{\sqrt{3}}$$

which leads consistently to

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

Thus the final answer from the solution conclusion and corrected working is Option A.