Why is the final answer $\frac{\pi}{2} + \tan^{-1}(\frac{1}{4})$ and not just $\tan^{-1}(\dots)$?

The angle $\angle AOB$ is the total opening between the two vectors. Since $OA$ is on the positive y-axis and $OB$ is in the third quadrant, the angle spans across the second quadrant. The $\frac{\pi}{2}$ represents the rotation from the y-axis to the negative x-axis, and $\tan^{-1}(\frac{1}{4})$ is the additional rotation into the third quadrant.

JEE Main 2026 (23 January Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: Let the line $y - x = 1$ intersect the ellipse $\frac{x ^{2}}{2} + \frac{y ^{2}}{1} = 1$ at the points $A$ and $B$ . Then the angle made by the line segment $A B$ at the center of the ellipse is :

Q: Why can't we just use the formula for the angle between two lines $\tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$ directly?

You absolutely can! However, remember that $m_1$ (the slope of $OA$) is undefined because $A$ lies on the y-axis ($x=0$). When a slope is undefined, the formula becomes tricky. It is often more intuitive to find the angle each vector makes with the axes and add them up, as we did here.

Q: How do I know which quadrant the point $B$ is in?

Look at the signs of the coordinates! We found $B = (-\frac{4}{3}, -\frac{1}{3})$. Since both $x$ and $y$ are negative, the point must reside in the third quadrant. Visualizing this helps you understand why we add the angle $\alpha$ to the $90^\circ$ segment.

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the geometry. The ellipse

\frac{x ^{2}}{2} + y^{2} = 1

is centered at the origin. The line

y = x + 1

cuts through it. Before diving into algebra, sketch the ellipse and the line. Where does the line hit the y-axis? That point is one of your intersection points.

Hint 2

Hint 2: You need the angle

∠ A O B

where

O

(0, 0)

. Instead of using the complex dot product formula, think about the slopes of the lines

O A

and

O B

. If

m_{1}

is the slope of

O A

and

m_{2}

is the slope of

O B

, the angle

θ

between them satisfies

tan θ = ∣ \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}} ∣

. Alternatively, find the individual angles each vector makes with the coordinate axes.

Hint 3

Hint 3: After solving the quadratic equation

\frac{3 x ^{2}}{2} + 2 x = 0

, you get

x_{1} = 0

and

x_{2} = - \frac{4}{3}

. Use these to find the coordinates

(x_{1}, y_{1})

and

(x_{2}, y_{2})

. Once you have the coordinates, calculate the angle of each vector relative to the axes using

tan θ = \frac{y}{x}

Visualized Solution

Visualizing the Ellipse

Ellipse Equation: $\frac{x ^{2}}{2} + \frac{y ^{2}}{1} = 1$
Center: $O (0, 0)$
Semi-major axis $a = 2$ , Semi-minor axis $b = 1$

Introducing the Intersecting Line

Line Equation: $y - x = 1 \Rightarrow y = x + 1$
Slope $m = 1$ , y-intercept $c = 1$

Strategy for Intersection Points

Goal: Find intersection points $A$ and $B$ .
Method: Substitute $y = x + 1$ into the ellipse equation.

Substitution and Expansion

Substitute: $\frac{x ^{2}}{2} + (x + 1)^{2} = 1$
Expand: $\frac{x ^{2}}{2} + (x^{2} + 2 x + 1) = 1$

Solving the Quadratic Equation

Simplify: $\frac{3 x ^{2}}{2} + 2 x = 0$
Factorize: $x (\frac{3 x}{2} + 2) = 0$
Roots: $x = 0$ and $x = - \frac{4}{3}$

Coordinates of Point $A$

For $x = 0$ , substitute in $y = x + 1$
$y = 0 + 1 = 1$
Point $A = (0, 1)$

Coordinates of Point $B$

For $x = - \frac{4}{3}$ , substitute in $y = x + 1$
$y = - \frac{4}{3} + 1 = - \frac{1}{3}$
Point $B = (- \frac{4}{3}, - \frac{1}{3})$

Visualizing Vectors $O A$ and $O B$

Origin $O (0, 0)$
Vector $O A$ connects center to $A$ .
Vector $O B$ connects center to $B$ .

Analyzing Vector $O A$

Point $A$ is on the positive y-axis.
Angle of $O A$ with positive x-axis is $9 0^{\circ}$ or $\frac{π}{2}$ .
Angle of $O A$ with negative x-axis is also $9 0^{\circ}$ or $\frac{π}{2}$ .

Analyzing Vector $O B$ and Angle $α$

Point $B (- \frac{4}{3}, - \frac{1}{3})$ is in the 3rd quadrant.
Let $α$ be the angle $O B$ makes with the negative x-axis.
$tan (α) = \frac{∣ y ∣}{∣ x ∣} = \frac{\frac{1}{3}}{\frac{4}{3}} = \frac{1}{4}$
$α = tan^{- 1} (\frac{1}{4})$

Final Angle Calculation

Total angle $∠ A O B = Angle of O A + Angle of O B$
$∠ A O B = \frac{π}{2} + α$
Final Answer: $\frac{π}{2} + tan^{- 1} (\frac{1}{4})$

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The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Solution Diagram

Analyzing the Setup

The ellipse is defined by the equation:

\frac{x ^{2}}{2} + \frac{y ^{2}}{1} = 1

We are interested in the angle subtended by the chord

A B

at the origin

O (0, 0)

, where the line

y = x + 1

intersects the ellipse.

The Intersection

To find the intersection points, we substitute

y = x + 1

into the ellipse equation:

\frac{x ^{2}}{2} + (x + 1)^{2} = 1

Expanding the squared term, we obtain:

\frac{x ^{2}}{2} + x^{2} + 2 x + 1 = 1

Simplifying this expression leads to:

\frac{3 x ^{2}}{2} + 2 x = 0

Factoring out

x

, we find the

x

-coordinates of the intersection points:

x (\frac{3 x}{2} + 2) = 0 \Rightarrow x_{1} = 0, x_{2} = - \frac{4}{3}

Mapping the Points

Using the line equation

y = x + 1

, we determine the corresponding

y

-coordinates for our points

A

and

B

For

x_{1} = 0

, we have

y_{1} = 0 + 1 = 1

. Thus, point

A

(0, 1)

For

x_{2} = - \frac{4}{3}

, we have

y_{2} = - \frac{4}{3} + 1 = - \frac{1}{3}

. Thus, point

B

(- \frac{4}{3}, - \frac{1}{3})

The Angular Perspective

Vector

O A

lies along the positive

y

-axis, making an angle of

9 0^{\circ}

(or

\frac{π}{2}

radians) with the positive

x

-axis.

Vector

O B

lies in the third quadrant. Let

α

be the angle that

O B

makes with the negative

x

-axis. Using the coordinates of

B

tan α = \frac{∣ y _{B} ∣}{∣ x _{B} ∣} = \frac{1/3}{4/3} = \frac{1}{4}

Therefore, the angle

α

tan^{- 1} (\frac{1}{4})

Final Calculation

The total angle

∠ A O B

is the sum of the angle from the positive

y

-axis to the negative

x

-axis (

9 0^{\circ}

) and the angle

α

from the negative

x

-axis to the vector

O B

The final angle is:

θ = \frac{π}{2} + tan^{- 1} (\frac{1}{4})

This result represents the precise angular span subtended by the segment

A B

at the origin.

Common Traps & FAQs

Wait, why do we reject the quadratic formula and just factorize? Is there a risk of missing a root?

Great observation! Factorization is just a shortcut here because the constant term in our quadratic

\frac{3 x ^{2}}{2} + 2 x = 0

is zero. Whenever you see a quadratic of the form

a x^{2} + b x = 0

, always factor out the

x

first. It is mathematically identical to the quadratic formula, just much faster and less prone to arithmetic errors!

Why can't we just use the formula for the angle between two lines

tan θ = ∣ \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}} ∣

directly?

You absolutely can! However, remember that

m_{1}

(the slope of

O A

) is undefined because

A

lies on the y-axis (

x = 0

). When a slope is undefined, the formula becomes tricky. It is often more intuitive to find the angle each vector makes with the axes and add them up, as we did here.

How do I know which quadrant the point

B

is in?

Look at the signs of the coordinates! We found

B = (- \frac{4}{3}, - \frac{1}{3})

. Since both

x

and

y

are negative, the point must reside in the third quadrant. Visualizing this helps you understand why we add the angle

α

to the

9 0^{\circ}

segment.

Is there a geometric property of ellipses I should have used instead of solving for points?

While there are advanced properties involving eccentric angles, for a problem like this, the algebraic approach is the most robust. JEE Advanced often tests your ability to bridge the gap between coordinate geometry and basic trigonometry. Don't overcomplicate it if the direct path is clear!

Why is the final answer

\frac{π}{2} + tan^{- 1} (\frac{1}{4})

and not just

tan^{- 1} (\dots)

The angle

∠ A O B

is the total opening between the two vectors. Since

O A

is on the positive y-axis and

O B

is in the third quadrant, the angle spans across the second quadrant. The

\frac{π}{2}

represents the rotation from the y-axis to the negative x-axis, and

tan^{- 1} (\frac{1}{4})

is the additional rotation into the third quadrant.

Animated Solution for Mathematics - Conic Sections: Let the line $y - x = 1$ intersect the ellipse $\frac{x ^{2}}{2} + \frac{y ^{2}}{1} = 1$ at the points $A$ and $B$ . Then the angle made by the line segment $A B$ at the center of the ellipse is :

Select Answer:

Visualized Solution

Visualizing the Ellipse

Introducing the Intersecting Line

Strategy for Intersection Points

Substitution and Expansion

Solving the Quadratic Equation

Coordinates of Point $A$

Coordinates of Point $B$

Visualizing Vectors $O A$ and $O B$

Analyzing Vector $O A$

Analyzing Vector $O B$ and Angle $α$

Final Angle Calculation

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Intersection

Mapping the Points

The Angular Perspective

Final Calculation

Similar Questions

An ellipse has eccentricity $\frac{1}{2}$ and one focus at the point $P (\frac{1}{2}, 1)$ . Its one directrix is the common tangent, nearer to the point $P$ , to the circle $x^{2} + y^{2} = 1$ and the hyperbola $x^{2} - y^{2} = 1$ . The equation of the ellipse, in the standard form, is.........

An ellipse is drawn by taking a diameter of the circle $(x - 1)^{2} + y^{2} = 1$ as its semi-minor axis and a diameter of the circle $x^{2} + (y - 2)^{2} = 4$ as semi-major axis. If the centre of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is:

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The line $y = x + 1$ meets the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{2} = 1$ at two points $P$ and $Q$ . If $r$ is the radius of the circle with $P Q$ as diameter then $(3 r)^{2}$ is equal to

Select Answer:

Visualized Solution

Visualizing the Ellipse

Introducing the Intersecting Line

Strategy for Intersection Points

Substitution and Expansion

Solving the Quadratic Equation

Coordinates of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Coordinates of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Visualizing Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OA and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OB

Analyzing Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OA

Analyzing Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OB and Angle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Final Angle Calculation

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Intersection

Mapping the Points

The Angular Perspective

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the line 3x−2y+12=0 intersects the parabola 4y=3x2 at the points A and B, then at the vertex of the parabola, the line segment AB subtends an angle equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The line y=x+1 meets the ellipse 4x2​+2y2​=1 at two points P and Q. If r is the radius of the circle with PQ as diameter then (3r)2 is equal to

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