A line passing through the point P(sqrt(5), sqrt(5)) intersects the ellipse (x2/36) + (y2/25) = 1 at A and B [cite…

Year	2026	2025	2024	Total
Questions	29	44	21	94

Q8 jee_main_2026_23_january_evening Family of Curves

If the points of intersection of the ellipses

x² + 2y² - 6x - 12y + 23 = 0

and

4x² + 2y² - 20x - 12y + 35 = 0

lie on a circle of radius

r

and centre

(a, b)

, then the value of

ab + 18r²

is

A. $53$ $53$
B. $51$ $51$
C. $52$ $52$
D. $55$ $55$

Solution

Related Formula

The equation of a family of curves passing through the intersection of two conics $S₁ = 0$ $S_1 = 0$ and $S₂ = 0$ $S_2 = 0$ is $S₁ + λ S₂ = 0$ $S_1 + \lambda S_2 = 0$. For this resulting curve to be a circle, the coefficient of $x²$ $x^2$ must equal the coefficient of $y²$ $y^2$, and the coefficient of the $xy$ $xy$ term must be zero.

Core Logic

Let the two ellipses be: $S₁ \equiv x² + 2y² - 6x - 12y + 23 = 0$ $S₂ \equiv 4x² + 2y² - 20x - 12y + 35 = 0$

Equation of the curve passing through their intersection is $S₁ + λ S₂ = 0$ $S_1 + \lambda S_2 = 0$:

(1 + 4λ)x² + (2 + 2λ)y² - (6 + 20λ)x - (12 + 12λ)y + (23 + 35λ) = 0

For this to represent a circle, coefficient of $x²$ $x^2$ = coefficient of $y²$ $y^2$:

1 + 4λ = 2 + 2λ 2λ = 1 λ = (1)/(2)

Step 1: Finding Circle Parameters

Substitute $λ = 1/2$ $\lambda = 1/2$ back into the family equation:

(1 + 2)x² + (2 + 1)y² - (6 + 10)x - (12 + 6)y + (23 + (35)/(2)) = 0

3x² + 3y² - 16x - 18y + (81)/(2) = 0

Dividing by 3 to write in standard form:

x² + y² - (16)/(3)x - 6y + (27)/(2) = 0

The centre $(a, b)$ $(a, b)$ is given by $(-g, -f)$ $(-g, -f)$:

a = (8)/(3), b = 3

The radius $r$ $r$ is given by $r² = g² + f² - c$ $r^2 = g^2 + f^2 - c$:

r² = ((-8)/(3))² + (-3)² - (27)/(2) = (64)/(9) + 9 - (27)/(2) = (128 + 162 - 243)/(18) = (47)/(18)

Step 2: Final Calculation

We need to find $ab + 18r²$ $ab + 18r^2$:

ab = ((8)/(3))(3) = 8

18r² = 18((47)/(18)) = 47

ab + 18r² = 8 + 47 = 55

Pattern Recognition

When intersection points of two $2nd$ $2^{\text{nd}}$ degree curves form a circle, apply $S₁ + λ S₂ = 0$ $S_1 + \lambda S_2 = 0$ immediately, forcing the necessary symmetric coefficients to extract $λ$ $\lambda$.

Chapter Mix

Class 11 Maths: Conic Sections Class 11 Maths: Circles

Q9 jee_main_2026_23_january_evening Properties of Hyperbola

Let

PQ

be a chord of the hyperbola

x²4- y²b²=1

, perpendicular to the x-axis such that

OPQ

is an equilateral triangle,

O

being the centre of the hyperbola. If the eccentricity of the hyperbola is

\sqrt(3)

, then the area of the triangle

OPQ

is:

A. $2\sqrt(3)$ $2\sqrt{3}$
B. $8\sqrt(3)5$ $\frac{8\sqrt{3}}{5}$
C. $(11)/(5)$ $\frac{11}{5}$
D. $(9)/(5)$ $\frac{9}{5}$

Solution

Related Formula

Eccentricity of a hyperbola $(x²)/(a²) - (y²)/(b²) = 1$ $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $e = \sqrt(1 + (b²)/(a²))$ $e = \sqrt{1 + \frac{b^2}{a^2}}$ Parametric coordinates on a hyperbola are $(a θ, b θ)$ $(a\sec\theta, b\tan\theta)$.

Core Logic

Properties of Hyperbola diagram for Q9 - JEE Main 2026 Evening

Given

e = \sqrt(3)

and

a² = 4

(

a=2

):

e² = 1 + (b²)/(4) 3 = 1 + (b²)/(4) b² = 8 b = 2\sqrt(2)

The equation is $(x²)/(4) - (y²)/(8) = 1$ $\frac{x^2}{4} - \frac{y^2}{8} = 1$.

Let the point $P$ $P$ on the hyperbola be $(2 θ, 2\sqrt(2) θ)$ $(2\sec\theta, 2\sqrt{2}\tan\theta)$. Since chord $PQ$ $PQ$ is perpendicular to the x-axis, the triangle $OPQ$ $OPQ$ has the x-axis as its altitude $OM$ $OM$. In equilateral $OPQ$ $\triangle OPQ$, half the vertex angle is $30^°$ $30^\circ$:

30^° = (PM)/(OM)

Step 1: Calculating Coordinates

1\sqrt(3) = 2\sqrt(2) θ2 θ = \sqrt(2) θ

θ = 1\sqrt(6)

Step 2: Area Calculation

The area of $OPQ = 2 \times ((1)/(2) \times OM \times PM) = OM \times PM$

Area = (2 θ) \times (2\sqrt(2) θ) = 4\sqrt(2) ( θ)/( ²θ)

Since $θ = 1\sqrt(6)$ $\sin\theta = \frac{1}{\sqrt{6}}$, we have $²θ = 1 - (1)/(6) = (5)/(6)$ .

Area = 4\sqrt(2) 1\sqrt(6)(5)/(6) = 4\sqrt(2) ( 1\sqrt(6))((6)/(5)) = 4\sqrt(2) ( \sqrt(6)5) = 4\sqrt(12)5 = 8\sqrt(3)5

Pattern Recognition

A perpendicular chord symmetric about the principal axis automatically splits into two right triangles. Utilizing the parametric coordinates directly defines the ratio of sides matching $30^°$ $\tan 30^\circ$.

Chapter Mix

Class 11 Maths: Hyperbola

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

A line passing through the point P(√(5), √(5))$P(\sqrt{5}, \sqrt{5})$ intersects the ellipse (x²)/(36) + (y²)/(25) = 1$\frac{x^2}{36} + \frac{y^2}{25} = 1$ at A$A$ and B$B$ [cite: 567] such that (PA) · (PB)$(PA) \cdot (PB)$ is maximum. Then 5(PA² + PB²)$5(PA^{2} + PB^{2})$ is equal to

Solution & Explanation

Related Formula

Core Logic

Step 1: Maximization Condition

Step 2: Distance Value Summation

Pattern Recognition

Evaluation Rubric / Model Answer

Chapter Mix

More Conic Sections Previous-Year Questions — Page 4

Solution

Related Formula

Core Logic

Step 1: Finding Circle Parameters

Step 2: Final Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating Coordinates

Step 2: Area Calculation

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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