JEE Main 2023 — 30 January, Morning Shift. Previous Year Question.
Problem
A straight line cuts off the intercepts $OA=a$ and $OB=b$ on the positive directions of the $x$-axis and $y$-axis respectively. If the perpendicular from the origin $O$ to this line makes an angle of $\dfrac{\pi}6$ with the positive direction of the $y$-axis, and the area of $\triangle OAB$ is $\dfrac{98\sqrt3}3$, then $a^2-b^2$ is equal to:
Key insight. Writing the line in normal form — in terms of the perpendicular distance $p$ from the origin and the angle that perpendicular makes — expresses both intercepts $a$ and $b$ directly in terms of the single unknown $p$, turning a two-variable geometry problem into a one-variable algebra problem.
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Approach
Since the perpendicular from the origin makes a known angle with the $y$-axis, it makes the complementary angle with the $x$-axis — this pins down the triangle formed by $O$, the foot of the perpendicular, and each intercept as a right triangle with a known angle. Express $a$ and $b$ in terms of the perpendicular distance $p$ using basic trigonometry, substitute into the given area to solve for $p$, then compute $a^2-b^2$.
Solution
Step 1 — Relate the perpendicular’s angle to each intercept
Let $p$ be the perpendicular distance from $O$ to the line, meeting it at foot $F$. Since the perpendicular makes an angle of $\dfrac{\pi}6$ with the positive $y$-axis, it makes an angle of $\dfrac{\pi}2-\dfrac{\pi}6=\dfrac{\pi}3$ with the positive $x$-axis.
In right triangle $OFA$ (right-angled at $F$), the angle at $O$ between $OA$ and $OF$ is $\dfrac{\pi}3$, so:
$$\cos\frac{\pi}3 = \frac{p}{a} \implies a = \frac{p}{\cos(\pi/3)} = \frac{p}{1/2} = 2p$$
Step 2 — Find b in terms of p
In right triangle $OFB$, the angle at $O$ between $OB$ and $OF$ is $\dfrac{\pi}6$, so:
$$\sin\frac{\pi}3 = \frac{p}{b} \implies b = \frac{p}{\sin(\pi/3)} = \frac{p}{\sqrt3/2} = \frac{2p}{\sqrt3}$$
(Equivalently, this follows from the angle at $O$ between $OB$ and $OF$ being $\dfrac{\pi}6$, and $\cos\left(\dfrac{\pi}6\right)=\dfrac{\sqrt3}2$ appearing in the same ratio.)
Step 3 — Substitute into the given area
$$\text{Area}(\triangle OAB) = \frac12\,ab = \frac12(2p)\left(\frac{2p}{\sqrt3}\right) = \frac{2p^2}{\sqrt3}$$
Setting this equal to the given area:
$$\frac{2p^2}{\sqrt3} = \frac{98\sqrt3}3$$
Step 4 — Solve for p
Multiplying both sides by $\sqrt3$:
$$2p^2 = \frac{98\sqrt3\cdot\sqrt3}3 = \frac{98\times3}3 = 98$$
$$p^2 = 49 \implies p=7$$
Step 5 — Compute a² − b²
$$a^2 = (2p)^2 = 4p^2, \qquad b^2 = \left(\frac{2p}{\sqrt3}\right)^2 = \frac{4p^2}3$$
$$a^2-b^2 = 4p^2-\frac{4p^2}3 = 4p^2\left(1-\frac13\right) = 4p^2\cdot\frac23 = \frac{8p^2}3$$
With $p^2=49$:
$$a^2-b^2 = \frac{8\times49}3 = \frac{392}3$$
Answer
$$\frac{392}3$$
Common mistakes
- Mixing up which angle goes with which intercept. The perpendicular makes $\dfrac{\pi}6$ with the $y$-axis and $\dfrac{\pi}3$ with the $x$-axis — using $\dfrac{\pi}6$ where $\dfrac{\pi}3$ belongs (or vice versa) swaps the roles of $a$ and $b$ and flips the sign of the final answer.
- Computing $a$ and $b$ from the area formula alone, without first finding $p$. With two unknowns ($a$ and $b$) but only one equation (the area), the system is underdetermined until it’s rewritten in terms of the single variable $p$ — trying to solve for $a$ and $b$ directly from the area alone doesn’t work.
- Arithmetic slips when rationalizing $\dfrac{98\sqrt3\cdot\sqrt3}3$. Since $\sqrt3\times\sqrt3=3$, this simplifies to $\dfrac{98\times3}3=98$ cleanly — forgetting that $\sqrt3\times\sqrt3=3$ (rather than $\sqrt3$ or $9$) is an easy place to go wrong.
Practise next
- The same line’s perpendicular from the origin still makes an angle of $\dfrac{\pi}6$ with the positive $y$-axis, but the area of $\triangle OAB$ is now $\dfrac{32\sqrt3}3$. Find $a^2-b^2$.
Show answer
$\dfrac{128}{3}$. The relations $a=2p$, $b=\dfrac{2p}{\sqrt3}$ and $a^2-b^2=\dfrac{8p^2}{3}$ don’t depend on the specific area — only $p$ changes.
Setting $\dfrac{2p^2}{\sqrt3}=\dfrac{32\sqrt3}3$ gives $2p^2=32$, so $p^2=16$. Then $a^2-b^2=\dfrac{8\times16}3=\dfrac{128}3$.
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