Let y2 = 12x the parabola and S be its focus. Let PQ be a focal chord of the parabola such that (SP) (SQ) = rac1474…

Year	2026	2025	2024	Total
Questions	22	38	16	76

Q1 jee_main_2026_23_january_morning Hyperbola

Let the domain of the function

f(x) = ₃ ₅ ₇(9x - x² - 13)

be the interval

(m, n)

. Let the hyperbola

(x²)/(a²) - (y²)/(b²) = 1

have eccentricity

(n)/(3)

and the length of the latus rectum

(8m)/(3)

. Then

b² - a²

is equal to:

A. $5$ $5$
B. $11$ $11$
C. $9$ $9$
D. $7$ $7$

Solution

Related Formula

e = \sqrt(1 + (b²)/(a²))

L.R. = (2b²)/(a)

Core Logic

For the domain of the given logarithmic function, the argument of the innermost logarithm must be strictly greater than 1 because of the nested logs:

₅( ₇(9x-x²-13))>0

\Rightarrow ₇(9x-x²-13) > 1

\Rightarrow 9x-x²-13 > 7

\Rightarrow x²-9x+20 < 0

\Rightarrow (x-4)(x-5) < 0

Thus, $4 < x < 5$ $4 < x < 5$. Therefore, the domain interval is $(4, 5)$ $(4, 5)$, yielding $m = 4$ $m = 4$ and $n = 5$ $n = 5$.

Step 1: Hyperbola Properties

Given eccentricity $e = (n)/(3) = (5)/(3)$ $e = \frac{n}{3} = \frac{5}{3}$:

e = \sqrt(1 + (b²)/(a²)) = (5)/(3) \Rightarrow (b²)/(a²) = (25)/(9) - 1 = (16)/(9)

\Rightarrow (b)/(a) = (4)/(3)

Given the length of the latus rectum is $(8m)/(3)$ $\frac{8m}{3}$:

(2b²)/(a) = (8(4))/(3) = (32)/(3)

\Rightarrow 2b((b)/(a)) = (32)/(3) \Rightarrow 2b((4)/(3)) = (32)/(3)

\Rightarrow 8b = 32 \Rightarrow b = 4

Since $(b)/(a) = (4)/(3)$ $\frac{b}{a} = \frac{4}{3}$, we get $a = 3$ $a = 3$.

Step 2: Final Calculation

We need to find $b² - a²$ $b^2 - a^2$:

b² - a² = (4)² - (3)² = 16 - 9 = 7

Pattern Recognition

Nested logarithmic domains require unpacking from the outside in: $ₐ(X) > 0 \Rightarrow X > 1$ $\log_a(X) > 0 \Rightarrow X > 1$. Linking function domains to coordinate geometry parameters is a standard JEE cross-topic pattern.

Chapter Mix

Class 11 Maths: Conic Sections Class 11 Maths: Relations and Functions

Q5 jee_main_2026_23_january_morning Ellipse

Let the line

y - x = 1

intersect the ellipse

(x²)/(2) + (y²)/(1) = 1

at the points

A

and

B

. Then the angle made by the line segment

AB

at the center of the ellipse is:

A. $π - ⁻¹((1)/(4))$
B. $(π)/(2) + ⁻¹((1)/(4))$
C. $(π)/(2) + 2 ⁻¹((1)/(4))$
D. $(π)/(2) - ⁻¹((1)/(4))$

Solution

Related Formula

θ = | (m₁ - m₂)/(1 + m₁ m₂) |

Core Logic

Find the intersection points of the line $y = x + 1$ $y = x + 1$ and the ellipse $(x²)/(2) + y² = 1$ $\frac{x^2}{2} + y^2 = 1$.

Ellipse diagram for Q5 - JEE Main 2026 Morning

Substitute

y = x + 1

into the ellipse equation:

(x²)/(2) + (x + 1)² = 1

x² + 2(x² + 2x + 1) = 2

3x² + 4x = 0 \Rightarrow x(3x + 4) = 0

This gives $x = 0$ $x = 0$ or $x = -(4)/(3)$ $x = -\frac{4}{3}$.

Step 1: Calculate Intersection Points

For $x = 0$ $x = 0$, $y = 1 \Rightarrow A(0, 1)$ $y = 1 \Rightarrow A(0, 1)$. For $x = -(4)/(3)$ $x = -\frac{4}{3}$, $y = -(4)/(3) + 1 = -(1)/(3) \Rightarrow B(-(4)/(3), -(1)/(3))$ .

Step 2: Find Angle at the Origin

Let $O(0,0)$ $O(0,0)$ be the center of the ellipse. The angle made by segment $AB$ $AB$ at $O$ $O$ is $∠ AOB$ $\angle AOB$. The slope of $OA$ $OA$ is $m₁ = (1 - 0)/(0 - 0) = \infty$ $m_1 = \frac{1 - 0}{0 - 0} = \infty$ (which means $OA$ $OA$ is along the y-axis, angle is $π/2$ $\pi/2$). The slope of $OB$ $OB$ is $m₂ = (-1/3 - 0)/(-4/3 - 0) = (1)/(4)$ . The angle of $OB$ $OB$ with the positive x-axis is $θ = ⁻¹((1)/(4))$ . The total angle $∠ AOB$ $\angle AOB$ is $(π)/(2) + θ = (π)/(2) + ⁻¹((1)/(4))$ .

Pattern Recognition

When solving line-conic intersection, explicit extraction of points $(A, B)$ $(A, B)$ is often simpler than using homogenization if the intersection yields simple rational or integer coordinates.

Chapter Mix

Class 11 Maths: Conic Sections Class 11 Maths: Straight Lines

Q19 jee_main_2026_23_january_evening Properties of Parabola

An equilateral triangle

OAB

is inscribed in the parabola

y²=4x

with the vertex

O

at the vertex of the parabola. Then the minimum distance of the circle having

AB

as a diameter from the origin is

A. $4(3 - \sqrt(3))$ $4(3 - \sqrt{3})$
B. $2(8 - 3\sqrt(3))$ $2(8 - 3\sqrt{3})$
C. $4(6 + \sqrt(3))$ $4(6 + \sqrt{3})$
D. $2(3 + \sqrt(3))$ $2(3 + \sqrt{3})$

Solution

Related Formula

Parametric form of $y² = 4ax$ $y^2 = 4ax$ is $(at², 2at)$ $(at^2, 2at)$. Minimum distance from origin to a circle with centre $C$ $C$ and radius $r$ $r$ is $|OC - r|$ $|OC - r|$.

Core Logic

Properties of Parabola diagram for Q19 - JEE Main 2026 Evening

For parabola

y² = 4x

,

a = 1

. Let the vertices of the equilateral triangle be

O(0,0), A(t², 2t), B(t², -2t)

due to symmetry across the x-axis. The slope of

OA

makes an angle of

30^°

with the x-axis.

mOA = (2t - 0)/(t² - 0) = (2)/(t)

30^° = (2)/(t) 1\sqrt(3) = (2)/(t) t = 2\sqrt(3)

Step 1: Defining the Circle

Vertices are $A((2\sqrt(3))², 2(2\sqrt(3))) = A(12, 4\sqrt(3))$ and $B(12, -4\sqrt(3))$ $B(12, -4\sqrt{3})$. The circle has diameter $AB$ $AB$. Therefore, its centre is the midpoint of $AB$ $AB$, which is $C(12, 0)$ $C(12, 0)$. The radius $R$ $R$ is half the length of $AB$ $AB$, so $R = 4\sqrt(3)$ $R = 4\sqrt{3}$.

The equation of the circle is:

(x - 12)² + y² = (4\sqrt(3))²

Step 2: Calculating Distance

The distance from the origin $O(0,0)$ $O(0,0)$ to the centre $C(12, 0)$ $C(12, 0)$ is $d = 12$ $d = 12$. The minimum distance from the origin to the circle is $|d - R|$ $|d - R|$:

= 12 - 4\sqrt(3) = 4(3 - \sqrt(3))

Pattern Recognition

Any polygon inscribed in a standard parabola symmetric across the axis can be defined completely by equating the slope of a vertex from the origin to the respective tangent of the split angle.

Chapter Mix

Class 11 Maths: Parabola Class 11 Maths: Circles

Q2 jee_main_2026_24_january_morning Circles and Centroid Locus

Let a circle of radius

4

pass through the origin

O

, the points

A(-\sqrt(3)a, 0)

and

B(0, -\sqrt(2)b)

, where

a

and

b

are real parameters and

ab \neq 0

. Then the locus of the centroid of

Δ OAB

is a circle of radius

Circles and Centroid Locus diagram for Q2 - JEE Main 2026 Morning — Graphical representation of the triangle OAB mapped on the Cartesian plane.

A. $(5)/(3)$ $\frac{5}{3}$
B. $(7)/(3)$ $\frac{7}{3}$
C. $(8)/(3)$ $\frac{8}{3}$
D. $(11)/(3)$ $\frac{11}{3}$

Solution

Related Formula

Centroid (h,k) = ( (x₁ + x₂ + x₃)/(3), (y₁ + y₂ + y₃)/(3) )

Core Logic

Coordinate geometry setup for Centroid Locus Q2 — Graphical representation of the triangle OAB mapped on the Cartesian plane.

Since

∠ AOB = 90^°

,

AB

is the diameter of the circle passing through

O, A, B

. Therefore, the diameter

AB = 2r = 8

.

Step 1: Distance and Coordinate Mapping

$AB² = 64$ $AB^2 = 64$

(-\sqrt(3)a - 0)² + (0 - (-\sqrt(2)b))² = 64

3a² + 2b² = 64

Step 2: Locus of Centroid

Let the centroid be $G(h,k)$ $G(h,k)$.

h = -\sqrt(3)a + 0 + 03 \Rightarrow a = -\sqrt(3)h

k = 0 - \sqrt(2)b + 03 \Rightarrow b = - 3\sqrt(2)k

Substituting $a$ $a$ and $b$ $b$ into the equation:

3(-\sqrt(3)h)² + 2(- 3\sqrt(2)k)² = 64

9h² + 9k² = 64

x² + y² = (64)/(9)

Step 3: Radius of Locus

This is a circle with $r² = (64)/(9)$ $r^2 = \frac{64}{9}$, so $r = (8)/(3)$ $r = \frac{8}{3}$.

Pattern Recognition

Whenever a circle passes through the origin and points on the axes, the segment joining the axes points is always the diameter. Centroid coordinates easily scale the locus.

Chapter Mix

Class 11 Maths: Conic Sections Class 11 Maths: Straight Lines

Q7 jee_main_2026_24_january_morning Ellipse Properties

Let each of the two ellipses

E₁: x²a²+ y²b²=1,(a>b)

and

E₂: x²A²+ y²B²=1,(A$E_{2}:\frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}=1,(A have eccentricity (4)/(5)$\frac{4}{5}$. Let the lengths of the latus recta of E₁$E_{1}$ and E₂$E_{2}$ be ₁$\ell_{1}$ and ₂$\ell_{2}$, respectively, such that 2 ₁²=9 ₂$2\ell_{1}^{2}=9\ell_{2}$. If the distance between the foci of E₁$E_{1}$ is 8$8$, then the distance between the foci of E₂$E_{2}$ is Graphical mapping defining major and minor axes orientation.

A. $(96)/(5)$ $\frac{96}{5}$
B. $(32)/(5)$ $\frac{32}{5}$
C. $(16)/(5)$ $\frac{16}{5}$
D. $(8)/(5)$ $\frac{8}{5}$

Solution

Related Formula

Eccentricity e = 1 - minor²major²

Distance between foci = 2a e (or 2Be if vertical)

Latus rectum length = 2minor²major

Core Logic

For $E₁$ $E_1$: $a>b \Rightarrow e = (4)/(5)$ $a>b \Rightarrow e = \frac{4}{5}$. Distance between foci $2ae = 8 \Rightarrow a(4/5) = 4 \Rightarrow a = 5$ .

b² = a²(1 - e²) = 25(1 - 16/25) = 9

$₁ = (2b²)/(a) = (18)/(5)$

Step 1: Analysing E2

For $E₂$ $E_2$: $A$A Vertical ellipse. e = (4)/(5)$e = \frac{4}{5}$.$

$A² = B²(1 - e²) = B²(1 - (16)/(25)) = (9)/(25)B² \Rightarrow A = (3)/(5)B$
$₂ = (2A²)/(B) = (2(9/25)B²)/(B) = (18)/(25)B$

Step 2: Linking Condition

Given $2 ₁² = 9 ₂$ $2\ell_1^2 = 9\ell_2$:
$2((18)/(5))² = 9((18)/(25)B)$ $2 \times (324)/(25) = (162)/(25)B$ $(648)/(25) = (162B)/(25) \Rightarrow B = 4$
Step 3: Finding Final Distance

Distance between foci of $E₂$ $E_2$ is $2Be$ $2Be$.
$= 2(4)((4)/(5)) = (32)/(5)$
Pattern Recognition

Notice the axis orientation shift: $E₁$ $E_1$ is horizontal, $E₂$ $E_2$ is vertical. Apply latus rectum formula $(2A²)/(B)$ $\frac{2A^2}{B}$ appropriately to avoid formula blind-spots.

Chapter Mix

Class 11 Maths: Conic Sections

Solution & Explanation

Related Formula

Core Logic

Step 1: Compute Endpoint Coordinate Bounds

Step 2: Derive Circle Equation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Conic Sections Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Step 1: Hyperbola Properties

Step 2: Final Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Intersection Points

Step 2: Find Angle at the Origin

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Defining the Circle

Step 2: Calculating Distance

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Distance and Coordinate Mapping

Step 2: Locus of Centroid

Step 3: Radius of Locus

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Analysing E2

Step 2: Linking Condition

Step 3: Finding Final Distance

Pattern Recognition

Chapter Mix

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