If the equation of the parabola with vertex V((3/2),3) and the directrix x+2y=0 is alpha x2+beta y2-gamma…

Year	2026	2025	2024	Total
Questions	29	44	21	94

Q3 jee_main_2026_28_january_evening Parabola and Triangles

Let

A

be the focus of the parabola

y² = 8x

. Let the line

y = mx + c

intersect the parabola at two distinct points

B

and

C

. If the centroid of the triangle

ABC

is

((7)/(3), (4)/(3))

, then

(BC)²

is equal to :

A. $41$ $41$
B. $80$ $80$
C. $89$ $89$
D. $32$ $32$

Solution

Related Formula

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

D² = (x₂-x₁)² + (y₂-y₁)²

Core Logic

Focus of $y² = 8x$ $y^2 = 8x$ is $A(2, 0)$ $A(2, 0)$ since $4a = 8 \Rightarrow a=2$ $4a = 8 \Rightarrow a=2$. Let points on parabola be $B(2t₁², 4t₁)$ $B(2t_1^2, 4t_1)$ and $C(2t₂², 4t₂)$ $C(2t_2^2, 4t_2)$. The centroid of $Δ ABC$ $\Delta ABC$ is given as $((7)/(3), (4)/(3))$ $\left(\frac{7}{3}, \frac{4}{3}\right)$. Equating coordinates:

(2 + 2t₁² + 2t₂²)/(3) = (7)/(3) \Rightarrow t₁² + t₂² = (5)/(2)

(0 + 4t₁ + 4t₂)/(3) = (4)/(3) \Rightarrow t₁ + t₂ = 1

Parabola points and centroid configuration

Execution

Square the sum equation:

(t₁ + t₂)² = t₁² + t₂² + 2t₁t₂

1 = (5)/(2) + 2t₁t₂ \Rightarrow 2t₁t₂ = -(3)/(2) \Rightarrow t₁t₂ = -(3)/(4)

Find $(t₁ - t₂)²$ $(t_1 - t_2)^2$:

(t₁ - t₂)² = (t₁ + t₂)² - 4t₁t₂ = 1 - 4(-(3)/(4)) = 4

Calculate distance squared for $BC$ $BC$:

(BC)² = (2t₁² - 2t₂²)² + (4t₁ - 4t₂)²

(BC)² = 4(t₁ - t₂)²(t₁ + t₂)² + 16(t₁ - t₂)²

(BC)² = 4(4)(1) + 16(4) = 16 + 64 = 80

Pattern Recognition

Using parametric coordinates $(at², 2at)$ $(at^2, 2at)$ systematically reduces algebraic complexity when determining intersections or triangle properties on a parabola.

Chapter Mix

Class 11 Maths: Conic Sections

Q9 jee_main_2026_28_january_evening Ellipse Parameters and Latus Rectum

An ellipse has its center at

(1,-2)

, one focus at

(3,-2)

and one vertex at

(5, - 2)

. Then the length of its latus rectum is :

A. $16\sqrt(3)$ $\frac{16}{\sqrt{3}}$
B. $6$ $6$
C. $4\sqrt(3)$ $4\sqrt{3}$
D. $6\sqrt(3)$ $6\sqrt{3}$

Solution

Related Formula

Latus Rectum (LR) = (2b²)/(a) = 2a(1-e²)

Core Logic

From the given coordinates on the major axis ( $y = -2$ $y = -2$): Center $C(1, -2)$ $C(1, -2)$, Focus $F₁(3, -2)$ $F_1(3, -2)$, Vertex $A₁(5, -2)$ $A_1(5, -2)$. Distance from center to vertex, $CA₁ = a = 5 - 1 = 4$ $CA_1 = a = 5 - 1 = 4$. Distance from center to focus, $CF₁ = ae = 3 - 1 = 2$ $CF_1 = ae = 3 - 1 = 2$.

Ellipse dimensions mapped to coordinates

Execution

Calculate eccentricity $e$ $e$:

ae = 2 \Rightarrow 4e = 2 \Rightarrow e = (1)/(2)

Use alternate formula for Latus Rectum:

LR = 2e((a)/(e) - ae) or directly 2a(1-e²)

LR = 2(4)(1 - (1)/(4)) = 8 \times (3)/(4) = 6

Pattern Recognition

Aligning focus, center, and vertex along a constant y-axis implies a standard shifted ellipse where absolute differences in x-coordinates yield standard parameters ( $a$ $a$ and $ae$ $ae$) directly.

Chapter Mix

Class 11 Maths: Conic Sections

Q10 jee_main_2026_28_january_evening Confocal Ellipse and Hyperbola

Let the ellipse

E: x²144 + y²169 = 1

and the hyperbola

H: x²16 - y²λ² = -1

have the same foci. If

e

and

L

respectively denote the eccentricity and the length of the latus rectum of

H

, then the value of

24(e + L)

is:

A. $296$ $296$
B. $126$ $126$
C. $148$ $148$
D. $67$ $67$

Solution

Related Formula

e = \sqrt(1 - (a²)/(b²)) (for vertical ellipse)

e = \sqrt(1 + (a²)/(b²)) (for conjugate hyperbola)

Core Logic

For Ellipse $E: (x²)/(144) + (y²)/(169) = 1$ $a² = 144, b² = 169$ $a^2 = 144, b^2 = 169$. Since $b > a$ $b > a$, the major axis is along the y-axis. Eccentricity $e' = \sqrt(1 - (144)/(169)) = \sqrt((25)/(169)) = (5)/(13)$ . Foci of ellipse = $(0, \pm be') = (0, \pm 13 \times (5)/(13)) = (0, \pm 5)$ .

Execution

For Hyperbola $H: (y²)/(λ²) - (x²)/(16) = 1$ Foci of conjugate hyperbola are $(0, \pm λ e)$ $(0, \pm \lambda e)$. Equating foci: $λ e = 5$ $\lambda e = 5$.

e = \sqrt(1 + (16)/(λ²))

λ \sqrt(1 + (16)/(λ²)) = 5 \Rightarrow λ² + 16 = 25 \Rightarrow λ² = 9 \Rightarrow λ = 3

Eccentricity of hyperbola, $e = (5)/(3)$ $e = \frac{5}{3}$. Length of latus rectum of hyperbola, $L = (2(16))/(λ) = (32)/(3)$ .

Calculate $24(e + L)$ $24(e + L)$:

24(e + L) = 24[(5)/(3) + (32)/(3)] = 24((37)/(3)) = 8 \times 37 = 296

Pattern Recognition

Confocal conics usually align along the same major axis. Notice the $-1$ $-1$ on the RHS of the hyperbola equation indicates a conjugate hyperbola orienting it vertically to match the $b>a$ $b>a$ ellipse.

Chapter Mix

Class 11 Maths: Conic Sections

Q12 jee_main_2026_28_january_evening Parametric Form and Chord Intersections

Let the circle

x² + y² = 4

intersect x-axis at the points

A(a, 0), a > 0

and

B(b, 0)

. Let

P(2 α, 2 α), 0 < α < (π)/(2)

and

Q(2 β, 2 β)

be two points such that

(α - β) = (π)/(2)

. Then the point of intersection of

AQ

and

BP

lies on:

A. $x² + y² - 4y - 4 = 0$ $x^{2} + y^{2} - 4y - 4 = 0$
B. $x² + y² - 4x - 4 = 0$ $x^{2} + y^{2} - 4x - 4 = 0$
C. $x² + y² - 4x - 4y = 0$ $x^{2} + y^{2} - 4x - 4y = 0$
D. $x² + y² - 4x - 4y - 4 = 0$ $x^{2} + y^{2} - 4x - 4y - 4 = 0$

Solution

Core Logic

Intersection of circle with x-axis provides $A(2,0)$ $A(2,0)$ and $B(-2,0)$ $B(-2,0)$. Let the point of intersection of $AQ$ $AQ$ and $BP$ $BP$ be $R(h, k)$ $R(h, k)$. Since $R$ $R$ lies on $BP$ $BP$, the slope $mBR = mBP$ $m_{BR} = m_{BP}$:

(k)/(h + 2) = (2 α)/(2 α + 2) = (α)/(2)

Since $R$ $R$ lies on $AQ$ $AQ$, the slope $mAR = mAQ$ $m_{AR} = m_{AQ}$:

(k)/(h - 2) = (2 β)/(2 β - 2) = ( β)/( β - 1) = - (β)/(2)

Execution

We are given $α - β = (π)/(2) \Rightarrow (α)/(2) - (β)/(2) = (π)/(4)$ . Applying the $(A-B)$ $\tan(A-B)$ formula:

((α)/(2) - (β)/(2)) = ( (α)/(2) - (β)/(2))/(1 + (α)/(2) (β)/(2)) = 1

Substitute the slope relations: $(α)/(2) = (k)/(h+2)$ $\tan\frac{\alpha}{2} = \frac{k}{h+2}$ $(β)/(2) = -(h-2)/(k)$ $\tan\frac{\beta}{2} = -\frac{h-2}{k}$ (since $- (β)/(2) = (k)/(h-2)$ $-\cot\frac{\beta}{2} = \frac{k}{h-2}$)

$1 = ((k)/(h+2) + (h-2)/(k))/(1 + ((k)/(h+2))((2-h)/(k)))$ $1 = k² + h² - 4(k(h+2))/(k) \cdot (k(h+2) - k(2-h))/(k(h+2)) wait, clear denominator$ $1 = (k² + h² - 4)/(k(h+2) + k(2-h)) \times k(h+2)$ The denominator simplifies to:

1 + (2-h)/(h+2) = (h+2+2-h)/(h+2) = (4)/(h+2)

Numerator is $(k² + h² - 4)/(k(h+2))$ $\frac{k^2 + h^2 - 4}{k(h+2)}$. So the expression simplifies to:

1 = (k² + h² - 4)/(4k)

h² + k² - 4k - 4 = 0

Locus of $R$ $R$ is $x² + y² - 4y - 4 = 0$ $x^2 + y^2 - 4y - 4 = 0$.

Pattern Recognition

Connecting chords from extreme diameter vertices to points whose parametric angles differ by $π/2$ $\pi/2$ reliably generates perpendicular-like slope products or standard tangent angle identities, mapping directly to a circular locus.

Chapter Mix

Class 11 Maths: Circles

Q55 jee_main_2025_02_april_evening Ellipse

If the length of the minor axis of an ellipse is equal to one fourth of the distance between the foci, then the eccentricity of the ellipse is :

A. $4\sqrt(17)$ $\frac{4}{\sqrt{17}}$
B. $\sqrt(3)16$ $\frac{\sqrt{3}}{16}$
C. $3\sqrt(19)$ $\frac{3}{\sqrt{19}}$
D. $\sqrt(5)7$ $\frac{\sqrt{5}}{7}$

Solution

Related Formula

Length of minor axis = 2b

Distance between foci = 2ae

Eccentricity: e = \sqrt(1 - (b²)/(a²))

Core Logic

We set up an algebraic equation relating $b$ $b$, $a$ $a$, and $e$ $e$ from the given geometric condition, then substitute it into the eccentricity identity.

Step 1: Set up the geometric relation

Given that $2b = (1)/(4) (2ae)$ $2b = \frac{1}{4} (2ae)$:

b = (ae)/(4) (b)/(a) = (e)/(4)

Square both sides:

(b²)/(a²) = (e²)/(16)

Step 2: Solve for eccentricity

Using the eccentricity relation:

e² = 1 - (b²)/(a²)

e² = 1 - (e²)/(16)

e² (1 + (1)/(16)) = 1

(17)/(16) e² = 1 e² = (16)/(17) e = 4\sqrt(17)

Pattern Recognition

Standard Ellipse relations: For standard ellipses, the ratio of axes and the eccentricity are coupled quadratic equations. Expressing $b/a$ $b/a$ as a function of $e$ $e$ allows direct solving of the eccentricity.

Chapter Mix

Class 11 Mathematics: Conic Sections

If the equation of the parabola with vertex V((3)/(2),3)$V\left(\frac{3}{2},3\right)$ and the directrix x+2y=0$x+2y=0$ is α x²+β y²-γ xy-30x-60y+225=0,$\alpha x^{2}+\beta y^{2}-\gamma xy-30x-60y+225=0,$ then α+β+γ$\alpha+\beta+\gamma$ is equal to:

Solution & Explanation

Related Formula

Step 1: Determine the Focus coordinates

Step 2: Derive the Parabola Locus Equation

Step 3: Coefficient Extraction

Pattern Recognition

Chapter Mix

Reference Study Guides

More Conic Sections Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Execution

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Execution

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Execution

Pattern Recognition

Chapter Mix

Solution

Core Logic

Execution

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Set up the geometric relation

Step 2: Solve for eccentricity

Pattern Recognition

Chapter Mix

More Conic Sections Questions — jee_main_2025_24_jan_evening

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back

If the equation of the parabola with vertex $V((3)/(2),3)$ $V\left(\frac{3}{2},3\right)$ and the directrix $x+2y=0$ $x+2y=0$ is $α x²+β y²-γ xy-30x-60y+225=0,$ then $α+β+γ$ $\alpha+\beta+\gamma$ is equal to: