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 $\Delta OAB$ is $\frac{98}{\sqrt{3}}$, then $a^2 - b^2$ is equal to: Options: (A) $\frac{392}{3}$ (B) 196 (C) $\frac{196}{3}$ (D) 98

### Analyzing the Setup Imagine you are standing at the origin of a Cartesian plane. You see a line slicing through the first quadrant, creating a right-angled triangle with the axes. When a problem provides the perpendicular distance from the origin, you should abandon the standard $y = mx + c$ form. Instead, we reach for the elegant, powerful **Normal Form**: $$x \cos \theta + y \sin \theta = p$$ ### The Angle Trap The problem states the perpendicular makes an angle of $\frac{\pi}{6}$ with the $y$-axis. However, the $\theta$ in the Normal Form is defined as the angle with the **positive x-axis**. This is a classic JEE trap. Since the $x$-axis and $y$-axis are at $\frac{\pi}{2}$, the angle with the $x$-axis must be: $$\theta = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$$ Substituting this into our equation, we get: $$x \cos\left(\frac{\pi}{3}\right) + y \sin\left(\frac{\pi}{3}\right) = p \Rightarrow \frac{x}{2} + \frac{\sqrt{3}y}{2} = p$$ ### The Bridge to Intercepts We need the intercepts $a$ and $b$. The intercept form of a line is $\frac{x}{a} + \frac{y}{b} = 1$. Let us manipulate our Normal Form equation to match this by dividing by $p$: $$\frac{x}{2p} + \frac{y}{\frac{2p}{\sqrt{3}}} = 1$$ By comparison, we identify the intercepts as: $$a = 2p, \quad b = \frac{2p}{\sqrt{3}}$$ ### The Final Convergence The area of $\Delta OAB$ is given as $\frac{98}{\sqrt{3}}$. Using the area formula for a right triangle, $\frac{1}{2}ab$, we substitute our expressions: $$\text{Area} = \frac{1}{2}(2p)\left(\frac{2p}{\sqrt{3}}\right) = \frac{2p^2}{\sqrt{3}}$$ Equating this to the given value: $$\frac{2p^2}{\sqrt{3}} = \frac{98}{\sqrt{3}} \Rightarrow 2p^2 = 98 \Rightarrow p^2 = 49$$ With $p = 7$, we calculate the intercepts: $$a = 14, \quad b = \frac{14}{\sqrt{3}}$$ Squaring these values yields $a^2 = 196$ and $b^2 = \frac{196}{3}$. Finally, we compute the result: $$a^2 - b^2 = 196 - \frac{196}{3} = 196\left(1 - \frac{1}{3}\right) = 196\left(\frac{2}{3}\right) = \mathbf{\frac{392}{3}}$$

JEE Main 2023 (30 January Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Straight Lines: A straight line cuts off the intercepts $O A = a$ and $O B = b$ on the positive directions of x-axis and y-axis respectively. If the perpendicular from origin O to this line makes an angle of $\frac{π}{6}$ with positive direction of y-axis and the area of $Δ O A B$ is $\frac{98}{3}$ , then $a^{2} - b^{2}$ is equal to:

Q: Wait, why is $\theta = \frac{\pi}{3}$? The question says $\frac{\pi}{6}$!

Great catch! The Normal Form $x \cos \theta + y \sin \theta = p$ defines $\theta$ as the angle the perpendicular makes with the *positive x-axis*. Since the perpendicular makes $\frac{\pi}{6}$ with the y-axis, and the axes are perpendicular ($90^\circ$), the angle with the x-axis is $90^\circ - 30^\circ = 60^\circ$, or $\frac{\pi}{3}$.

Q: Is it possible for $p$ to be negative?

In the context of the Normal Form $x \cos \theta + y \sin \theta = p$, $p$ represents the perpendicular distance from the origin to the line. Distance is a magnitude, so it must be positive. If you ever get a negative value for $p$, check your geometry or your sign conventions!

Select Answer:

Visualized Solution

Visualizing the Line and Intercepts

Let the line intersect the x-axis at $A (a, 0)$ and the y-axis at $B (0, b)$ .
The intercepts are $O A = a$ and $O B = b$ .

The Normal Form of a Line

The normal form of a straight line is given by $x cos θ + y sin θ = p$ ,
where $p$ is the perpendicular distance from the origin,
and $θ$ is the angle it makes with the positive x-axis.

Finding the Angle $θ$

The perpendicular makes an angle of $\frac{π}{6}$ with the positive y-axis.
The angle with the positive x-axis is $θ = \frac{π}{2} - \frac{π}{6} = \frac{π}{3}$ .

Substituting $θ$ into the Equation

Substitute $θ = \frac{π}{3}$ into the normal form:
$x cos (\frac{π}{3}) + y sin (\frac{π}{3}) = p$
$⟹ \frac{x}{2} + \frac{3 y}{2} = p$

Finding Intercepts $a$ and $b$

Rearrange into intercept form $\frac{x}{a} + \frac{y}{b} = 1$ .
Divide by $p$ : $\frac{x}{2 p} + \frac{y}{\frac{2 p}{3}} = 1$ .
Thus, $a = 2 p$ and $b = \frac{2 p}{3}$ .

Using the Area Condition

The area of $Δ O A B$ is $\frac{1}{2} ab$ .
Substitute $a$ and $b$ : $Area = \frac{1}{2} (2 p) (\frac{2 p}{3}) = \frac{2 p ^{2}}{3}$ .

Equating with Given Area

We are given that the area is $\frac{98}{3}$ .
Equating the two expressions: $\frac{2 p ^{2}}{3} = \frac{98}{3}$ .

Solving for $p$

Cancel $3$ from both sides: $2 p^{2} = 98 ⟹ p^{2} = 49$ .
Since distance $p > 0$ , we get $p = 7$ .

Calculating $a^{2}$ and $b^{2}$

Recall $a = 2 p ⟹ a = 14 ⟹ a^{2} = 196$ .
And $b = \frac{2 p}{3} ⟹ b = \frac{14}{3} ⟹ b^{2} = \frac{196}{3}$ .

Final Calculation of $a^{2} - b^{2}$

We need to find $a^{2} - b^{2}$ .
Substitute the values: $196 - \frac{196}{3}$
$= 196 (1 - \frac{1}{3}) = 196 (\frac{2}{3}) = \frac{392}{3}$ .

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The Sigma Insight: Various Forms of Equations of a Line

Solution Diagram

Analyzing the Setup

Imagine you are standing at the origin of a Cartesian plane. You see a line slicing through the first quadrant, creating a right-angled triangle with the axes.

When a problem provides the perpendicular distance from the origin, you should abandon the standard

y = m x + c

form. Instead, we reach for the elegant, powerful Normal Form:

x cos θ + y sin θ = p

The Angle Trap

The problem states the perpendicular makes an angle of

\frac{π}{6}

with the

y

-axis. However, the

θ

in the Normal Form is defined as the angle with the positive x-axis.

This is a classic JEE trap. Since the

x

-axis and

y

-axis are at

\frac{π}{2}

, the angle with the

x

-axis must be:

θ = \frac{π}{2} - \frac{π}{6} = \frac{π}{3}

Substituting this into our equation, we get:

x cos (\frac{π}{3}) + y sin (\frac{π}{3}) = p \Rightarrow \frac{x}{2} + \frac{3 y}{2} = p

The Bridge to Intercepts

We need the intercepts

a

and

b

. The intercept form of a line is

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

Let us manipulate our Normal Form equation to match this by dividing by

p

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

By comparison, we identify the intercepts as:

a = 2 p, b = \frac{2 p}{3}

The Final Convergence

The area of

Δ O A B

is given as

\frac{98}{3}

. Using the area formula for a right triangle,

\frac{1}{2} ab

, we substitute our expressions:

Area = \frac{1}{2} (2 p) (\frac{2 p}{3}) = \frac{2 p ^{2}}{3}

Equating this to the given value:

\frac{2 p ^{2}}{3} = \frac{98}{3} \Rightarrow 2 p^{2} = 98 \Rightarrow p^{2} = 49

With

p = 7

, we calculate the intercepts:

a = 14, b = \frac{14}{3}

Squaring these values yields

a^{2} = 196

and

b^{2} = \frac{196}{3}

. Finally, we compute the result:

a^{2} - b^{2} = 196 - \frac{196}{3} = 196 (1 - \frac{1}{3}) = 196 (\frac{2}{3}) = \frac{392}{3}

Common Traps & FAQs

Wait, why is

θ = \frac{π}{3}

? The question says

\frac{π}{6}

Great catch! The Normal Form

x cos θ + y sin θ = p

defines

θ

as the angle the perpendicular makes with the *positive x-axis*. Since the perpendicular makes

\frac{π}{6}

with the y-axis, and the axes are perpendicular (

9 0^{\circ}

), the angle with the x-axis is

9 0^{\circ} - 3 0^{\circ} = 6 0^{\circ}

, or

\frac{π}{3}

Why can't we just use

y = m x + c

You absolutely could, but it would be a nightmare of algebra! You'd have to relate

m

and

c

to the perpendicular distance

p

using complex formulas. The Normal Form is the 'shortcut' designed specifically for problems involving perpendicular distances from the origin. Always choose the tool that fits the data!

Is it possible for

p

to be negative?

In the context of the Normal Form

x cos θ + y sin θ = p

p

represents the perpendicular distance from the origin to the line. Distance is a magnitude, so it must be positive. If you ever get a negative value for

p

, check your geometry or your sign conventions!

Select Answer:

Visualized Solution

Visualizing the Line and Intercepts

The Normal Form of a Line

Finding the Angle $θ$

Substituting $θ$ into the Equation

Finding Intercepts $a$ and $b$

Using the Area Condition

Equating with Given Area

Solving for $p$

Calculating $a^{2}$ and $b^{2}$

Final Calculation of $a^{2} - b^{2}$

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Angle Trap

The Bridge to Intercepts

The Final Convergence

Similar Questions

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Select Answer:

Visualized Solution

Visualizing the Line and Intercepts

The Normal Form of a Line

Finding the Angle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ into the Equation

Finding Intercepts .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Using the Area Condition

Equating with Given Area

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2

Final Calculation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2−b2

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Angle Trap

The Bridge to Intercepts

The Final Convergence

Similar Questions

Finding the Angle $θ$

Substituting $θ$ into the Equation

Finding Intercepts $a$ and $b$

Solving for $p$

Calculating $a^{2}$ and $b^{2}$

Final Calculation of $a^{2} - b^{2}$