JEE Advanced Previous Year Questions (2018 - 2025): Conic Sections

Table of Contents
1. 2023
2. 2022
3. 2021
4. 2020
5. 2019
6. 2018
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Q1: A normal with slope 1/√6 is drawn from the point (0, -α) to the parabola x² = -4ay, where a > 0.
Let L be the line passing through (0, -α) and parallel to the directrix of the parabola.
Suppose that L intersects the parabola at two points A and B.
Let r denote the length of the latus rectum and s denote the square of the length of the line segment AB.
If r : s = 1 : 16, then the value of 24a is ______.     [JEE Advanced 2024 Paper 2]
Ans: 12

Slope of normal = 1t = 1√6 ⇒ t = √6
Now, -at² + α2at = 1t
⇒ -at² + α = 2a
⇒ -6a + α = 2a ⇒ α = 8a
For A and B:
x² = -4a(-8a)
⇒ x² = 32a² ⇒ x = ±4√2a
∴ A(-4√2a, -8a), B(4√2a, -8a)
∴ AB² = (8√2a)² = 128a² = s
Length of LR = r = 4a
∴ rs = 4a128a² = 116
∴ 32a = 16 ⇒ a = 12
∴ 24a = 12 Ans.

Q2: Let A₁, B₁, C₁ be three points in the xy-plane. Suppose that the lines A₁C₁ and B₁C₁ are tangents to the curve y² = 8x at A₁ and B₁, respectively. If O = (0,0) and C₁ = (-4,0), then which of the following statements is (are) TRUE?
(a) The length of the line segment OA₁ is 4√3
(b) The length of the line segment A₁B₁ is 16
(c) The orthocenter of the triangle A₁B₁C₁ is (0,0)
(d) The orthocenter of the triangle A₁B₁C₁ is (1,0)     [JEE Advanced 2024 Paper 2]
Ans: (a), (c)
Equation of tangent at (2t², 4t) is
ty = x + 2t²
Since it is passing through (-4, 0)
0 = -4 + 2t² ⇒ t = ±√2
Thus,
A₁ = (4, 4√2)
B₁ = (4, -4√2)
OA₁ = √48 = 4√3
A₁B₁ = 8√2
Equation of altitude of ΔA₁B₁C₁ drawn from A₁ is
y - 4√2 = √2(x - 4)
⇒ √2x - y = 0 ...(1)
Equation of altitude of ΔA₁B₁C₁ drawn from C₁ is
x = 0 ...(2)
Solving (1) and (2) ⇒ orthocenter is (0,0).
Correct options are (A), (C)

Q3: Consider the ellipse x²9 + y²4 = 1. Let S(p, q) be a point in the first quadrant such that p²9 + q²4 > 1. Two tangents are drawn from S to the ellipse, of which one meets the ellipse at one endpoint of the minor axis and the other meets the ellipse at a point T in the fourth quadrant. Let R be the vertex of the ellipse with positive x-coordinate and O be the center of the ellipse. If the area of the triangle △ORT is 32, then which of the following options is correct?
(a) q = 2, p = 3√3
(b) q = 2, p = 4√3
(c) q = 1, p = 5√3
(d) q = 1, p = 6√3     [JEE Advanced 2024 Paper 1]
Ans: (a)
Ar(△ORT) = 32

|12 × 3 × 2 sinθ| = 32
sinθ = 12 ⇒ θ = 11π6
T (3√32 , -1)
Tangent at (0,2),
x(0)9 + y(2)4 = 1 ⇒ y = 2 .........(1)
Tangent at (3√32 , -1)
x9 + y(-1)4 = 1 .........(2)
∴ By solving (1) & (2) ⇒ p = 3√3, q = 2
⇒ Option (A) is Correct.

2023

Q1: Let P be a point on the parabola y² = 4ax, where a > 0. The normal to the parabola at P meets the x-axis at a point Q. The area of the triangle PFQ, where F is the focus of the parabola, is 120 . If the slope m of the normal and a are both positive integers, then the pair (a, m) is  
(a) (2,3)
(b) (1,3)
(c) (2,4)
(d) (3,4)        [JEE Advanced 2023 Paper 1]
Ans: (a)

y² = 4ax
Equation of normal
y = mx - 2am - am³
Point of contact

and Point

Area of ΔPFQ =

a = 2, m = 3
Satisfies the equation (1), hence (2,3) will be the correct answer. 

Q2: Let T₁ and T₂ be two distinct common tangents to the ellipse  and the parabola . Suppose that the tangent T₁ touches P and E at the points A₁ and A₂, respectively and the tangent T₂ touches P and E at the points A₄ and A₃, respectively. Then which of the following statements is(are) true?
(a) The area of the quadrilateral A₁A₂A₃A₄ is 35 square units
(b) The area of the quadrilateral A₁A₂A₃A₄ is 36 square units
(c) The tangents T₁ and T₂ meet the x-axis at the point (-3, 0)
(d) The tangents T₁ and T₂ meet the x-axis at the point (-6, 0) [JEE Advanced 2023 Paper 1]
Ans: (a) and (c)

For common tangent

⇒ Equation of common tangents y =  x + 3 and y = -x - 3 point of contact for parabola is

Let A₂(x₁, y₁)⇒ tangent to E is =

A₃ is mirror image of A₂ in x-axis  ⇒ A₃ (-2, -1)

Intersection point of T₁ = 0 and T₂ = 0 is (-3, 0)

Area of quadrilateral A₁A₂A₃A₄ = 12(12 + 2) × 5 = 35 square units 

2022

Q1: Consider the hyperbola with foci at S and S₁, where S lies on the positive x-axis. Let P be a point on the hyperbola, in the first quadrant. Let ∠SPS₁ = α, with α < π/2. The straight line passing through the point S and having the same slope as that of the tangent at P to the hyperbola, intersects the straight line S₁P at S₁. Let δ be the distance of P from the straight line SP₁, and β = S₁P Then the greatest integer less than or equal to   is ________. [JEE Advanced 2022 Paper 2]
Ans: 

From property we know, tangent and normal is bisector of the angle between focal radii.

∴ Tangent AB divides the angle ∠SPS₁= α equal parts.
From another property, we know, if we draw perpendicular to the tangent on the hyperbola from two foci, then product of length of the perpendicular from foci = b²
∴ l × δ = b²
Given hyperbola,
∴ a² = 100
and b² = 64
∴ l × δ = 64 ....... (1)
From right angle triangle S₁ MP we get, sin⁡α/2 = l/β

Putting value of l in equation (1), we get

∴ Greatest integer = [7.1] = 7

Q2: Consider the ellipse  Let H(α, 0), 0 < α < 2, be a point. A straight line drawn through H parallel to the Y-axis crosses the ellipse and its auxiliary circle at points E and F respectively, in the first quadrant. The tangent to the ellipse at the point E intersects the positive x-axis at a point G. Suppose the straight line joining F and the origin makes an angle ϕ with the positive x-axis.

Area of triangle FGH =

=

Q3: Consider the parabola y² = 4x. Let S be the focus of the parabola. A pair of tangents drawn to the parabola from the point P = (-2, 1) meet the parabola at P₁ and P₂. Let Q₁ and Q₂ be points on the lines SP₁ and SP₂ respectively such that PQ1 is perpendicular to SP₁ and PQ₂ is perpendicular to SP₂. Then, which of the following is/are TRUE?
(a) SQ₁ = 2
(b)
(c) PQ₁ = 3
(d) SQ₂ = 1 [JEE Advanced 2022 Paper 1]
Ans: (b), (c) & (d)
Let P₁(t², 2t) then tangent at P₁ will be
ty = x + t²

Since, it passes through (-2,1)

So, we get point P₁(4, 4) and P₂(1, -2)
Now finding the equation of SP₁ : 4x - 3y - 4 = 0
And equation of SP₂: x - 1= 0
Now finding the point by foot of point on line formula, 

We get,

Now using the distance formula we get,

Hence, option (B, C, D) are correct.

2021

Q1: Let E be the ellipse . For any three distinct points P, Q and Q' on E, let M(P, Q) be the mid-point of the line segment joining P and Q, and M(P, Q') be the mid-point of the line segment joining P and Q'. Then the maximum possible value of the distance between M(P, Q) and M(P, Q'), as P, Q and Q' vary on E, is _______. [JEE Advanced 2021 Paper 2]
Ans: 4
As we know that, in a triangle, sides joining the mid-points of two sides is half and parallel to the third side.

Maximum value of QQ' is AA'
Hence, maximum value of
= 4

Q2: Let E denote the parabola y² = 8x. Let P = (-2, 4), and let Q and Q' be two distinct points on E such that the lines PQ and PQ' are tangents to E. Let F be the focus of E. Then which of the following statements is(are) TRUE?
(a) The triangle PFQ is a right-angled triangle
(b) The triangle QPQ' is a right-angled triangle
(c) The distance between P and F is 5√2
(d) F lies on the line joining Q and Q'                       [JEE Advanced 2021 Paper 2]
Ans: (a), (b) & (d)
Given, E : y² = 8x .... (i)
and P ≡ (-2, 4)
Now, directrix of Eq. (i) is x = -2 

So, point P(-2, 4) lies on the directrix of parabola y² = 8x. Hence,  (by the definition of director circle) and chord QQ' is a focal chord and segment PQ subtends a right angle at the focus.
Slope of PF = -1 (∵ PF ⊥ QQ')
Now, slope of

2020

Q1: Let a, b and λ be positive real numbers. Suppose P is an end point of the latus return of the
parabola y2 = 4λx, and suppose the ellipse  passes through the point P. If the tangents to the parabola and the ellipse at the point P are perpendicular to each other, then the eccentricity of the ellipse is
(a) 1/√2
(b) 1/2
(c) 1/3
(d) 2/5                     [JEE Advanced 2020 Paper 1]
Ans: (a)
Equation of given parabola is
y2 = 4λx ....(i)
So the end point of the latus rectum of the parabola (i), P(λ, 2λ) and the given ellipse , passes through point P(λ, 2λ).
On differentiating the equation of parabola, w.r.t. 'x', we get

∴ Slope of tangent to the parabola at point P is m₁ = 1
Similarly, on differentiating the equation of given ellipse, , w.r.t.x, we get

∴ Slope of tangent to the ellipse at point P is m₂ =
∵ It is given that the tangents are perpendicular to each other. So,  m₁m₂ = -1

∴ Eccentricity of ellipse  will be

Q2: Let a and b be positive real numbers such that a > 1 and b < a. Let P be a point in the first quadrant that lies on the hyperbola . Suppose the tangent to the hyperbola at P passes through the point (1, 0), and suppose the normal to the hyperbola at P cuts off equal intercepts on the coordinate axes. Let Δ denote the area of the triangle formed by the tangent at P, the normal at P and the X-axis. If e denotes the eccentricity of the hyperbola, then which of the following statements is/are TRUE?
(a) 1 < e < √2
(b) √2 < e < 2
(c) Δ = a⁴
(d) Δ = b⁴ [JEE Advanced 2020 Paper 2]
Ans: (a) & (d)
Equation of given hyperbola is
, a > b and a > 1
Let point P(a secθ, b tanθ) on the hyperbola in first quadrant i.e., .
Now equation of tangent to the hyperbola at point P is 

∵ The tangent (i) passes through point A(1, 0)
So,

and equation of normal to the hyperbola at point P having slope is , as normal cuts off equal intercepts on the coordinate axes, so slope must be -1. 

Therefore,  = -1

∵ The eccentricity of hyperbola

And the area of required triangle is, which is isosceles is 

=

2019

Q1: Let the circles C₁ : x² + y² = 9 and C₂ : (x - 3)² + (y - 4)² = 16, intersect at the points X and Y. Suppose that another circle C₃ : (x - h)² + (y - k)² = r² satisfies the following conditions :
(i) Centre of C₃ is collinear with the centres of C₁ and C₂.
(ii) C₁ and C₂ both lie inside C₃ and
(iii) C₃ touches C₁ at M and C₂ at N.
Let the line through X and Y intersect C₃ at Z and W, and let a common tangent of C₁ and C₃ be a tangent to the parabola x² = 8αy.
There are some expression given in the List-I whose values are given in List-II below. 

Which of the following is the only INCORRECT combination?
(a) (III), (R)
(b) (IV), (S)
(c) (I), (P)
(d) (IV), (U)                     [JEE Advanced 2019 Paper 2]
Ans: (b)

It is given that, the centres of circles C₁, C₂ and C₃ are co-linear, 

and MN is the length of diameter of circle C₃, so

= 3 + 5 + 4 = 12
So, radius of circle C₃, r = 6 ......(ii)
Since, the circle C₃ touches C₁ at M and C₂ at N, so
|C₁ C₃| = |r - 3|

From Eqs. (i) and (iii), we get

So,

Now, equation common chord XY of circles C₁ and C₂ is 

C₁ - C₂ = 0
⇒ 6x + 8y = 18
⇒ 3x + 4y = 9 ....(iv)
Now, PY² = GY² - GP²

Similarly, equation of ZW is 3x + 4y = 9.
Now, So, length of perpendicular from C₃ to

So,

{∵ C₃W = r = 6}

Now, area of

and area of ΔZMW = 1/2(ZW) (MP) 

∵ Common tangent of circles C₁ and C₃ is C₁ - C₃ = 0

∵ Tangent (v) is also touches the parabola x² = 8αy, 

So combination (iv), (S) is only incorrect.
Hence, option (b) is correct.

2018

Q1: Let , where a > b > 0, be a hyperbola in the XY-plane whose conjugate axis LM subtends an angle of 60^∘ at one of its vertices N. Let the area of the ΔLMN be 4√3. 

It is given,

and Area of ΔLMN = 4√3
Now, ΔLMN is an equilateral triangle whose sides is 2b .
Area of ΔLMN =

Also, area of  ΔLMN = 1/2 a(2b) = ab

(P) Length of conjugate axis = 2b = 2(2) = 4 (Q)

(R) Distance between the foci

(S) The length of latusrectum

P → 4; Q → 3; R → 1; S → 2
Hence, option (b) is correct. 

Q2: Let S be the circle in the XY-plane defined the equation x² + y² = 4.
Let E₁E₂ and F₁F₂ be the chords of S passing through the point P₀ (1, 1) and parallel to the X-axis and the Y-axis, respectively. Let G₁G₂ be the chord of S passing through P₀ and having slope-1. Let the tangents to S at E₁ and E₂ meet at E₃, then tangents to S at F₁ and F₂ meet at F₃, and the tangents to S at G₁ and G₂ meet at G₃. Then, the points E₃, F₃ and G₃ lie on the curve
(a) x + y = 4
(b) (x - 4)² + (y - 4)² = 16
(c) (x - 4)(y - 4) = 4
(d) xy = 4                  [JEE Advanced 2018 Paper 1]
Ans: (a)

Equation of tangent at  is

Intersection point of tangent at E₁ and E₂ is (0, 4)
∴ Coordinates of E₃ is (0, 4)
Similarly, equation of tangent at

x + √3y = 4, respectively and intersection point
is (4, 0), i.e., F₃(4, 0) and equation of tangent at G₁(0, 2) and G₂(2, 0) are 2y = 4 and 2x = 4, respectively and intersection point
is (2, 2) i.e., G₃(2, 2).
Point E₃(0, 4), F₃(4, 0) and G₃(2, 2) satisfies the line x + y = 4. 

Q3: Let S be the circle in the XY-plane defined the equation x² + y² = 4.
Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect the coordinate axes at the points M and N. Then, the mid-point of the line segment MN must lie on the curve
(a) (x + y)2 = 3xy
(b) x^2/3 + y^2/3 = 2^4/3
(c) x² + y² = 2xy
(d) x² + y² = x²y²           [JEE Advanced 2018 Paper 1]
Ans: (d)
We have,
x² + y² = 4
Let P(2 cosθ, 2 sinθ) be a point on a circle.
∴ Tangent at P is
2 cosθx + 2 sinθy = 4
⇒ x cosθ + y sinθ = 2 

 The coordinates at 
Let (h, k) is mid-point of MN 

∴ Mid-point of MN lie on the curve