JEE Main Previous Year Questions (2021-23): Conic Sections PDF Download

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Q.1. Let a tangent to the curve y² = 24x meet the curve xy = 2 at the points ⁢ and B. Then the mid points of such line segments AB lie on a parabola with the       (JEE Main 2023)
(a) Length of latus rectum 3/2
(b) directrix⁡ 4x = −3
(c) length of latus rectum 2
(d) directrix⁡ 4x = 3

Ans. d
c₁ : y² = 24 x  &  c² : xy = 2

AB : [Tangent to parabola at p(t)]
ty = x + 6t² …..(1)
AB : [chord with given mid point of hyperbola]

Q.2. Let a tangent to the curve 9x² + 16y² = 144 intersect the coordinate axes at the points ⁢ and B. Then, the minimum length of the line segment ⁢B is       (JEE Main 2023)

Ans. 7

Q.3. The equations of the sides ⁢B and ⁢C of a triangle ⁢BC are (λ + 1)x + λy = 4 and λx + (1 − λ)y + λ = 0 respectively. Its vertex ⁢ is on the y - axis and its orthocentre is (1,2). The length of the tangent from the point C to the part of the parabola y² = 6x in the first quadrant is :      (JEE Main 2023)
(a) 4
(b) 2
(c) √6
(d) 2√2

Ans. d

Q.4. 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.

The correct option is:    (JEE Advanced 2022)
(a) (I) → (R); (II) → (S); (III) → (Q); (IV) → (P)
(b) (I) → (R); (II) → (T); (III) → (S); (IV) → (P)
(c) (I) →(Q); (II) → (T); (III) → (S); (IV) → (P)
(d) (I) → (Q); (II) → (S); (III) → (Q); (IV) → (P)

Ans. c

Q.5. 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?     (JEE Advanced 2022)
(a) SQ₁ = 2
(b) Q₁Q₂ = (3√10)/5
(c) PQ₁ = 3
(d) SQ₂ = 1

Ans. b, c and d

Q.6. 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 P₁. 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)

Ans. 7

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,

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


= 64/9 = 7.1
∴ Greatest integer = [7.1] = 7

Q.7. Let the focal chord of the parabola P : y² = 4x along the line L : y = mx + c,m > 0 meet the parabola at the points M and N. Let the line L be a tangent to the hyperbola H : x² − y² =4. If O is the vertex of P and F is the focus of H on the positive x-axis, then the area of the quadrilateral OMFN is :     (JEE Main 2022)
(a) 2√6
(b) 1√4
(c) 4√6
(d) 4√14

Ans. b


Focus (ae, 0)
F(2√2, 0)
y = mx + c passes through (1, 0)
0 = m + C ...... (i)
L is tangent to hyperbola


m² = 4m² - 4
m = 2√3
C = -2/√3

P : y² = 4x


Area



= √56
= 2√14

Q.8. Let a line L pass through the point of intersection of the lines bx + 10y − 8 = 0 and 2x − 3y = 0, b ∈ R − {4/3}. If the line L also passes through the point (1,1) and touches the circle 17(x² + y²) = 16, then the eccentricity of the ellipse  is :     (JEE Main 2022)
(a) 2/√5
(b)
(c) 1/√5
(d)

Ans. b
L₁ : bx + 10y − 8 = 0, L₂ : 2x − 3y = 0
then L : (bx + 10y − 8) + λ(2x − 3y) = 0
∵ It passes through (1,1)
∴ b+2−λ=0⇒λ=b+2
and touches the circle x² + y² = (16/17)

⇒ 4λ² + b² + 4bλ + 100 + 9λ² − 60λ = 68
⇒ 13(b+2)² + b² + 4b(b+2) − 60(b+2) + 32 = 0
⇒ 18b² = 36
∴ b² = 2
∴ Eccentricity of ellipse :  is

Q.9. Let the hyperbola  pass through the point (2√2, −2√2). A parabola is drawn whose focus is same as the focus of H with positive abscissa and the directrix of the parabola passes through the other focus of H. If the length of the latus rectum of the parabola is e times the length of the latus rectum of H, where e is the eccentricity of H, then which of the following points lies on the parabola?      (JEE Main 2022)
(a) (2√3, 3√2)
(b) (3√3, −6√2)
(c) (√3, −√6)
(d) (3√6, 6√2)

Ans. b

Focus of parabola : (ae,0)
Directrix : x = −ae.
Equation of parabola ≡ y² = 4aex
Length of latus rectum of parabola = 4ae
Length of latus rectum of hyperbola = 2.b²/a
as given, 4ae = (2b²/a). e
2 = b²/a² ....(i)
∵ H passes through (2√2, −2√2) ⇒ 8/(a²) − 8/(b²) = 1 ........ (ii)
From (i) and (ii) a² = 4 and b² = 8 ⇒ e = 3
⇒ Equation of parabola is y² = 8√3x.

Q.10. If the tangents drawn at the points P and Q on the parabola y² = 2x − 3 intersect at the point R(0,1), then the orthocentre of the triangle PQR is :     (JEE Main 2022)
(a) (0, 1)
(b) (2, -1)
(c) (6, 3)
(d) (2, 1)

Ans. b

Equation of chord PQ
⇒ y × 1 = x − 3
⇒ x − y = 3
For point P & Q
Intersection of PQ with parabola P:(6, 3) Q:(2, −1)
Slope of RQ = −1 & Slope of PQ = 1
Therefore ∠PQR = 90^∘ ⇒ Orthocentre is at Q : (2, -1)

Q.11. If the length of the latus rectum of a parabola, whose focus is (a, a) and the tangent at its vertex is x + y = a, is 16, then |a| is equal to :     (JEE Main 2022)
(a) 2√2
(b) 2√3
(c) 4√2
(d) 4

Ans. c
Equation of tangent at vertex : L ≡ x + y − a = 0
Focus : F ≡ (a, a)
Perpendicular distance of L from F

Length of latus rectum
Given
⇒ |a| = 4√2

Q.12. Let P(a, b) be a point on the parabola y² = 8x such that the tangent at P passes through the centre of the circle x² + y² − 10x − 14y + 65 = 0. Let A be the product of all possible values of a and B be the product of all possible values of b. Then the value of A+B is equal to    (JEE Main 2022)
(a) 0
(b) 25
(c) 40
(d) 65

Ans. d
Centre of circle x² + y² − 10x − 14y + 65 = 0 is at (5, 7).
Let the equation of tangent to y2=8x is
yt = x + 2t²
which passes through (5, 7)
7t = 5 + 2t²
⇒ 2t² − 7t + 5 = 0
t = 1, (5/2)
A = 2 × 1² × 2 × (5/2)² = 25
B = 2 × 2 × 1 × 2 × 2 × (5/2) = 40
A + B = 65

Q.13. If the line x − 1 = 0 is a directrix of the hyperbola kx² − y² = 6, then the hyperbola passes through the point.     (JEE Main 2022)
(a) (−2√5, 6)
(b) (−√5, 3)
(c) (√5, −2)
(d) (2√5, 3√6)

Ans. c
Given hyperbola :
Eccentricity = e =
Directrices :
As given :
⇒ k = 2
Here hyperbola is
Checking the option gives (√5,−2) satisfies it.

Q.14. The equation of a common tangent to the parabolas y = x² and y = −(x − 2)² is     (JEE Main 2022)
(a) y = 4(x - 2)
(b) y = 4(x - 1)
(c) y = 4(x + 1)
(d) y = 4(x + 2)

Ans. b
Equation of tangent of slope m to y = x2
y = mx − (1/4)m²
Equation of tangent of slope m to y= −(x − 2)²
y = m(x − 2) + (1/4)m²
If both equation represent the same line
(1/4)m² − 2m = −(1/4)m²
m = 0, 4
So, equation of tangent
y = 4x − 4

Q.15. The acute angle between the pair of tangents drawn to the ellipse 2x² + 3y² = 5 from the point (1, 3) is     (JEE Main 2022)
(a)
(b)
(c)
(d)

Ans. b
2x² + 3y² = 5
Equation of tangent having slope m.

which passes through (1, 3)


Acute angle between the tangents is given by

Q.16. Let P and Q be any points on the curves (x−1)² + (y+1)² = 1 and y = x², respectively. The distance between P and Q is minimum for some value of the abscissa of P in the interval     (JEE Main 2022)
(a) (0, (1/4))
(b) ((1/2), (3/4))
(c) ((1/4), (1/2))
(d) ((3/4), 1)

Ans. c
y = mx + 2a + (1/(m²)) (Equation of normal to x² = 4ay in slope form) through (1,−1).
4m³ + 6m² + 1 = 0
⇒ m ≃ −1.6
Slope of normal ≃ (−8)/5 = tan⁡θ

Q.17. Let the tangent drawn to the parabola y² = 24x at the point (α, β) is perpendicular to the line 2x + 2y =5. Then the normal to the hyperbola  at the point (α + 4, β + 4) does NOT pass through the point:     (JEE Main 2022)
(a) 25, 10
(b) 20, 12
(c) 30, 8
(d) 15, 13

Ans. d
Any tangent to y² = 24x at (α, β)
βy = 12(x + α)
Slope = (12/β) and perpendicular to 2x + 2y = 5
⇒(12/β) = 1 ⇒ β = 12, α = 6
Hence hyperbola is (x²/36) - (y²/144) = 1and normal is drawn at (10, 16)
Equation of normal

This does not pass though (15, 13) out of given option.

Q.18. Let the foci of the ellipse  and the hyperbola  coincide. Then the length of the latus rectum of the hyperbola is:     (JEE Main 2022)
(a) 32/9
(b) 18/5
(c) 27/4
(d) 27/10

Ans. d


If foci coincide then
Hence, hyperbola is
Length of latus rectum

Q.19. The tangents at the points A(1,3) and B(1,−1) on the parabola y² − 2x − 2y = 1 meet at the point P. Then the area (in unit 2 ) of the triangle PAB is:     (JEE Main 2022)
(a) 4
(b) 6
(c) 7
(d) 8

Ans. d
Given curve : y² − 2x − 2y = 1.
Can be written as
(y − 1)2 = 2(x + 1)

And, the given information can be plotted as shown in figure
Tangent at A: 2y − x − 5 = 0 {using T =0}
Intersection with y = 1 is x = −3
Hence, point P is (−3,1)
Taking advantage of symmetry
Area of ΔPAB = 2 × (1/2) × (1−(−3)) × (3−1) = 8 sq. units

Q.20. If the ellipse  meets the line  on the x-axis and the line  on the y-axis, then the eccentricity of the ellipse is     (JEE Main 2022)
(a) 5/7
(b)
(c) 3/7
(d)

Ans. a
meets the line  on the x-axis
So, a = 7 and  meets the line  on the y-axis
So, b = 2√6
Therefore, e² = 1 - (b²/a²) = 1 - (24/49)
e = 5/7

Q.21. Let the eccentricity of the ellipse x² + a²y² = 25a² be b times the eccentricity of the hyperbola x² − a²y² = 5, where a is the minimum distance between the curves y = e^x and y = log_ex. Then a² + (1/b²) is equal to :    (JEE Main 2022)
(a) 3/2
(b) 5/2
(c) 3
(d) 5

Ans. d
Given ellipse
eccentricity

Also, given hyperbola,
x² - a²y² = 5

eccentricity

Also given,
e₁ = b × e₂


y = log_e^x is inverse of y = e^x so it is mirror image of each other with respect to y = x line.
Slope of tangent to y = e^x curve
dy/dx = e^x
Slope of tangent to  curve,

Both tangents are parallel to y = x line for minimum distance condition.
∴ Slope of y = x line = Slope of both the tangent.

∴ dy/dx = e^x = 1 ⇒ e^x = e⁰ = x = 0
∴ y = e^x = e⁰ = 1
and dy/dx =1/x = 1 ⇒ x = 1
∴ y  =
∴ tangent at (0, 1) point of y = e^x curve and tangent at (1, 0) point of y = log_e^x curve are parallel.
∴ Minimum distance between point (0, 1) and (1, 0) is


So,  

Q.22. Let x² + y² + Ax + By + C = 0 be a circle passing through (0, 6) and touching the parabola y = x² at (2, 4). Then A + C is equal to ______.      (JEE Main 2022)
(a) 16
(b) 88/5
(c) 72
(d) -8

Ans. a
For tangent to parabola y = x² at (2, 4)

Equation of tangent is y − 4 = 4(x−2)
⇒ 4x − y − 4 = 0
Family of circle can be given by
(x−2)² + (y−4)² + λ(4x−y−4) = 0
As it passes through (0,6)
2² + 2² + λ(−10) = 0
⇒ λ = 4/5
Equation of circle is



So, A + C = 16

Q.23. Let the maximum area of the triangle that can be inscribed in the ellipse  having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the y-axis, be 6√3. Then the eccentricity of the ellipse is :     (JEE Main 2022)
(a)
(b) 1/2
(c)
(d)

Ans. a
Given ellipse

∴ Let A(θ) be the area of ΔABB'
Then A(θ) = (1/2)4sin⁡θ (a + acos⁡θ)
A′(θ) = a(2cos⁡θ + 2cos²θ)
For maxima A′(θ) = 0
⇒ cos⁡θ = 1, cos⁡θ = (1/2)
But for maximum area cos⁡θ = (1/2)
∴ A⁡(θ) = 6√3

⇒ a = 4

Q.24. A particle is moving in the xy-plane along a curve C passing through the point (3, 3). The tangent to the curve C at the point P meets the x-axis at Q. If the y-axis bisects the segment PQ, then C is a parabola with     (JEE Main 2022)
(a) length of latus rectum 3
(b) length of latus rectum 6
(c) focus ((4/3), 0)
(d) focus (0, (3/4))

Ans. a
According to the question (Let P(x, y))
2x - (y(dx/dy)) = 0
(∵ equation of tangent at P : y − y = (dy/dx)(y−x))

⇒ 2ln⁡y = ln⁡ x + ln⁡ c
⇒ y² = cx
∵  this curve passes through (3, 3)
∴ c = 3
∴ required parabola y² = 3x and L.R = 3

Q.25. Let x = 2t, y = t²/3 be a conic. Let S be the focus and B be the point on the axis of the conic such that SA ⊥ BA, where A is any point on the conic. If k is the ordinate of the centroid of the ΔSAB, then  is equal to :    (JEE Main 2022)
(a) 17/18
(b) 19/18
(c) 11/18
(d) 13/18

Ans. d
x = 2t, y = (2/3)
t→1 A ≡ (2,(1/3))
Given conic is x² = 12y ⇒ S ≡ (0, 3)
Let B ≡ (0, β)
Given SA ⊥ BA

Q.26. If y = m₁x + c₁ and y = m₂x + c₂, m₁ ≠ m₂ are two common tangents of circle x² + y² = 2 and parabola y² = x, then the value of 8|m₁m₂| is equal to :     (JEE Main 2022)
(a) 3 + 4√2
(b) −5 + 6√2
(c) −4 + 3√2
(d) 7 + 6√2

Ans. c
Let tangent to y² = x be
y = mx + (1/4m)
For it being tangent to circle.

⇒ 32m⁴ + 32m² - 1  = 0

⇒ 8m₁m₂ = -4 + 3√2

Q.27. The line y = x + 1 meets the ellipse  at two points P and Q. If r is the radius of the circle with PQ as diameter then (3r)² is equal to:     (JEE Main 2022)
(a) 20
(b) 12
(c) 11
(d) 8

Ans. a
Let point (a, a + 1) as the point of intersection of line and ellipse.
So,  1 ⇒ a² + 2(a² + 2a + 1) = 4
⇒ 3a² + 4a − 2 = 0
If roots of this equation are α and β.
So, P(α, α + 1) and Q(β, β + 1)
PQ = 4r² = (α − β)² + (α − β)²

= (1/2)[16 + 24] = 20

Q.28. If the line y= 4 + kx, k > 0, is the tangent to the parabola y = x − x² at the point P and V is the vertex of the parabola, then the slope of the line through P and V is :     (JEE Main 2022)
(a) 3/2
(b) 26/9
(c) 5/2
(d) 23/6

Ans. c
∵ Line y = kx + 4 touches the parabola y = x − x².
So, kx + 4 = x − x² ⇒ x² + (k − 1)x + 4 = 0 has only one root
(k−1)² = 16 ⇒ k = 5 or −3 but k > 0
So, k = 5.
And hence x² + 4x + 4 = 0 ⇒ x = −2
So, P(−2, −6) and V is ((1/2), (2/4))
Slope of

Q.29. The normal to the hyperbola  at the point (8, 3√3) on it passes through the point :     (JEE Main 2022)
(a) (15, -2√3)
(b) (9, 2√3)
(c) (-1, 9√3)
(d) (-1, 6√3)

Ans. c
Given hyperbola :
∵ It passes through (8, 3√3)

Now, equation of normal to hyperbola

⇒ 2x + √3y = 25 ...... (i)
(−1, 9√3) satisfies (i)

Q.30. The locus of the mid point of the line segment joining the point (4, 3) and the points on the ellipse x² + 2y² = 4 is an ellipse with eccentricity :     (JEE Main 2022)
(a) √3/2
(b) 1/2√2
(c) 1/√2
(d) 1/2

Ans. c
Let P(2cos⁡θ, √2sin⁡θ) be any point on ellipse (x²/4) + (y²/2) =1 and Q(4, 3) and let (h, k) be the mid point of PQ then h =

Q.31. If m is the slope of a common tangent to the curves (x²/16) + (y²/9) = 1 and x² + y² =12, then 12m² is equal to :     (JEE Main 2022)
(a) 6
(b) 9
(c) 10
(d) 12

Ans. b

Let  be any tangent to C₁ and if this is also tangent to C₂ then

⇒ 16m² + 9 = 12m² + 12
⇒ 4m² = 3 ⇒ 12m² = 9

Q.32. Let the normal at the point on the parabola y² = 6x pass through the point (5, −8). If the tangent at P to the parabola intersects its directrix at the point Q, then the ordinate of the point Q is :     (JEE Main 2022)
(a) -3
(b) -(9/4)
(c) -(5/2)
(d) -2

Ans. b
Let P(at², 2at) where a = 3/2
T : yt = x + at² So point Q is (−a, at − (a/t))
N : y = −tx + 2at + at³ passes through (5, −8)
−8 = −5t + 3t + (3/2)t³
⇒ 3t³ − 4t + 16 = 0
⇒ (t + 2) (3t² − 6t + 8) = 0
⇒ t = 2
So ordinate of point Q is −(9/4).

Q.33. Let the eccentricity of an ellipse x²a² + y²b² = 1, a > b, be 1/4. If this ellipse passes through the point , then a² + b² is equal to :      (JEE Main 2022)
(a) 29
(b) 31
(c) 32
(d) 34

Ans. b


a²(1 − e²) = b²
a²(1 − 116) = b²
15a² = 16b² ⇒ a² = 16b²/15
From (i)

a² + b² = 15 + 16 = 31

Q.34. If the equation of the parabola, whose vertex is at (5, 4) and the directrix is 3x + y − 29 = 0, is x² + ay² + bxy + cx + dy + k = 0, then a + b + c + d + k is equal to     (JEE Main 2022)
(a) 575
(b) -575
(c) 576
(d) -576

Ans. d
Given vertex is (5, 4) and directrix 3x + y − 29 = 0
Let foot of perpendicular of (5, 4) on directrix is (x₁, y₁)

∴ (x₁, y₁) ≡ (8, 5)
So, focus of parabola will be S = (2, 3)
Let P(x, y) be any point on parabola, then

⇒ 10(x² + y² − 4x − 6y + 13) = 9x² + y² + 841 + 6xy − 58y − 174x
⇒ x² + 9y² − 6xy + 134x − 2y−711 = 0
and given parabola
x² + ay² + bxy + cx+ dy + k = 0
∴ a = 9, b = −6, c = 134, d = −2, k = −711

Q.35. Let the eccentricity of the hyperbola  and length of its latus rectum be 6√2. If y = 2x + c is a tangent to the hyperbola H, then the value of c² is equal to     (JEE Main 2022)
(a) 18
(b) 20
(c) 24
(d) 32

Ans. b

c² = a²m² - b² = 8.4 - 12 = 20

Q.36. If vertex of a parabola is (2, −1) and the equation of its directrix is 4x − 3y = 21, then the length of its latus rectum is :     (JEE Main 2022)
(a) 2
(b) 8
(c) 12
(d) 16

Ans. b
Vertex of Parabola : (2, −1) and directrix : 4x − 3y = 21
Distance of vertex from the directrix

∴ length of latus rectum = 4a = 8

Q.37. Let a > 0, b > 0. Let e and l respectively be the eccentricity and length of the latus rectum of the hyperbola  Let e' and l' respectively be the eccentricity and length of the latus rectum of its conjugate hyperbola. If  then the value of 77a + 44b is equal to :      (JEE Main 2022)
(a) 100
(b) 110
(c) 120
(d) 130

Ans. d

and e^′2 = 118l′ (l' be the length of LR of conjugate hyperbola)
⇒ a² + b² = (11/4)a²b ...(ii)
By (i) and (ii)
7a = 4b
then by (i)

⇒ 44b = 65 and 77a = 65
∴ 77a + 44b = 130

Q.38. Let P : y² = 4ax, a > 0 be a parabola with focus S. Let the tangents to the parabola P make an angle of (π/4) with the line y = 3x + 5 touch the parabola P at A and B. Then the value of a for which A, B and S are collinear is    (JEE Main 2022)
(a) 8 only
(b) 2 only
(c) 14 only
(d) any a > 0

Ans. d




⇒ m − 3 = 1 + 3m and m − 3 = −1 − 3m
⇒ 2m = −4 and 4m = 2
m = −2 and m = (1/2)
We know, Equation of tangent to the parabola y² = 4m is y = mx + (a/m) and point of contact is
∴ Equation of tangent
y = −2x − (a/2) and y = (x/2) + 2a
∴ Point of contact A and B are

As points A, B and S are colinear so area of triangle formed by those 3 points are zero.
Area of ΔABS =
= (a/4)(4a−0) + a(4a−a) + 1(0 − 4a²)
= a² + 3a² − 4a² = 0
∴ Area of triangle is independent of value of a.
So, for all value of a > 0 (already given a must be greater than 0) point A, B and S will be collinear.

Q.39. Let PQ be a focal chord of the parabola y² = 4x such that it subtends an angle of (π/2) at the point (3, 0). Let the line segment PQ be also a focal chord of the ellipse  a²>b². If e is the eccentricity of the ellipse E, then the value of 1/(e²) is equal to:      (JEE Main 2022)
(a) 1 + √2
(b) 3 + 2√2
(c) 1 + 2√3
(d) 4 + 5√3

Ans. b



⇒ t = ±1
∴ P ≡ (1, 2) & Q(1, -2)
∴ for ellipse (1/a²) + (4/b²) = 1 and ae = 1

Q.40. Two tangent lines l₁ and l₂ are drawn from the point (2, 0) to the parabola 2y² = −x. If the lines l₁ and l₂ are also tangent to the circle (x−5)² +y² = r, then 17r is equal to ___________.      (JEE Main 2022)

Ans. 9



4y = x−2 and 4y + x = 2
If these are also tangent to circle then d_c = r

⇒ 17r = 17 . (9/17) = 9

Q.41. Let the tangents at the points P and Q on the ellipse  meet at the point R(√2, 2√2 − 2). If S is the focus of the ellipse on its negative major axis, then SP² + SQ² is equal to  _____.     (JEE Main 2022)

Ans. 13

⇒ −2√2m(2√2−2) = 4 − 12 + 8√2
⇒ −4√2m(√2 − 1) = 8((√2) − 1)
⇒ m = − √2 and m → ∞
∴ Tangents are x = √2 and y = −√2x + √8
∴ P(√2, 0) and Q(1, √2) and S=(0, −√2)
∴ (PS)² + (QS)² = 4 + 9 = 13

Q.42. For the hyperbola H : x² − y² = 1 and the ellipse  a > b > 0, let the
(1) eccentricity of E be reciprocal of the eccentricity of H, and
(2) the line  be a common tangent of E and H. Then 4(a² + b²) is equal to  ______.    (JEE Main 2022)

Ans. 3
The equation of tangent to hyperbola x² − y² = 1 within slope m is equal to
And for same slope m, equation of tangent to ellipse
∵ Equation (i) and (ii) are identical
∴ a²m² + b² = m²−1
∴ m² =
But equation of common tangent is

∴ 5a² + 2b² = 3 ....... (i)
eccentricity of ellipse = 1/√2

⇒ a² = 2b² ....... (ii)
From equation (i) and (ii) : a² = (1/2), b² = (1/4)
∴ 4(a² + b²⁾ = 3

Q.43. A common tangent T to the curves  and 

 does not pass through the fourth quadrant. If T touches C₁ at (x₁, y₁) and C₂ at (x₂, y₂), then |2x₁ + x₂| is equal to _______.    (JEE Main 2022)

Ans. 20
Equation of tangent to ellipse  and given slope m is :
For slope m equation of tangent to hyperbola is :

Tangents from (i) and (ii) are identical then
4m2+9=42m2−143
∴ m = ±2 (+2 is not acceptable)
∴ m = −2.
Hence, x₁ = 8/5 and x₂ = 84/5

Q.44. If the length of the latus rectum of the ellipse x² + 4y² + 2x + 8y − λ = 0 is 4 , and l is the length of its major axis, then λ + l is equal to ______.     (JEE Main 2022)

Ans. 75
Equation of ellipse is : x² + 4y² + 2x + 8y − λ = 0
(x + 1)² + 4(y + 1)² = λ + 5

Length of latus rectum
∴ λ = 59.
Length of major axis
∴ λ + l = 75.

Q.45. An ellipse  passes through the vertices of the hyperbola  Let the major and minor axes of the ellipse E coincide with the transverse and conjugate axes of the hyperbola H, respectively. Let the product of the eccentricities of E and H be (1/2). If l is the length of the latus rectum of the ellipse E, then the value of 113l is equal to _____.     (JEE Main 2022)

Ans. 1552
Vertices of hyperbola = (0, ±8)
As ellipse pass through it i.e.,
0 + (64/b²) = 1 ⇒ b² = 64 ...... (1)
As major axis of ellipse coincide with transverse axis of hyperbola we have b > a i.e.



 


L.R of ellipse
= l = 1552/113
∴ 113l = 1552

Q.46. The sum of diameters of the circles that touch (i) the parabola 75x² = 64(5y−3) at the point ((8/5), (6/5)) and (ii) the y-axis, is equal to _______.     (JEE Main 2022)

Ans. 10

Equation of tangent to the parabola at

⇒ 120x = 160y
⇒ 3x = 4y
Equation of circle touching the given parabola at P can be taken as

If this circle touches y-axis then


⇒ D = 0


Radius =1 or 4
Sum of diameter = 10

Q.47. Let the equation of two diameters of a circle x² + y² − 2x + 2fy + 1 = 0 be 2px − y = 1 and 2x + py = 4p. Then the slope m ∈ (0, ∞) of the tangent to the hyperbola 3x² − y² = 3 passing through the centre of the circle is equal to _______.     (JEE Main 2022)

Ans. 2

2p + f − 1 = 0…(1)
2 − pf − 4p = 0…(2)
2 = p(f + 4)
p = 2/(f + 4)
2p = 1 − f
4/(f + 4) = 1 − f
f² + 3f = 0
f = 0 or −3
Hyperbola 3x² − y² = 3, x² − (y²/3) = 1
 
It passes (1, 0)

m tends ∞
It passes (1, 3)

(3 - m)² = m² - 3
m = 2

Q.48. Let PQ be a focal chord of length 6.25 units of the parabola y² = 4x. If O is the vertex of the parabola, then 10 times the area (in sq. units) of ΔPOQ is equal to ______.     (JEE Main 2022)

Ans. 25

Given parabola y² = 4x
∴ a = 1
Here, P, S, Q points are collinear.
∴ Slope of PS = Slope of QS






⇒ (t₂ − t₁)(t₁t₂ + 1) = 0
As t₂ − t₁ ≠ 0
∴ t₁t₂ + 1 = 0
t₁t₂ = −1
Now, lenght of PQ



= (t₁ - t₂)²
Given, length of PQ = (t₁ - t₂)² = 6.25
⇒ t1 - t2 = 2.5
Now, Area of ΔOPQ


= |−1×2.5|
= 2.5
∴ 10ΔOPQ = 10 x (25/10)= 25

Q.49. If two tangents drawn from a point (α, β) lying on the ellipse 25x² + 4y² = 1 to the parabola y² = 4x are such that the slope of one tangent is four times the other, then the value of (10α + 5)² + (16β² + 50)² equals _____.     (JEE Main 2022)

Ans. 2929
∵ (α, β) lies on the given ellipse, 25α² + 4β² = 1 ...(i)
Tangent to the parabola, y = mx + (1/m) passes through (α, β). So, αm² − βm + 1 = 0 has roots m₁ and 4m₁,

m₁ + 4m₁ = (β/α) and m₁⋅4m₁ = 1/α
Gives that 4β² = 25α ...(ii)
from (i) and (ii)
25(α² + α) = 1 ...(iii)
Now, (10α+5)² + (16β²+50)²
= 25(2α+1)² + 2500(2α+1)²
= 2525(4α² + 4α + 1) from equation (iii)
= 2525((4/25) + 1)
= 2929

Q.50. Let P₁ be a parabola with vertex (3, 2) and focus (4, 4) and P₂ be its mirror image with respect to the line x + 2y = 6. Then the directrix of P₂ is x + 2y = _____.    (JEE Main 2022)

Ans. 10
Focus = (4, 4) and vertex = (3, 2)
∴ Point of intersection of directrix with axis of parabola = A = (2, 0)
Image of A(2, 0) with respect to line x + 2y = 6 is B(x₂, y₂)


Point B is point of intersection of direction with axes of parabola P₂.
∴ x + 2y = λ must have point ((18/5), (16/5))
∴ x +2y = 10

Q.51. Let the hyperbola H : (x²/a²) − y² = 1 and the ellipse E : 3x² + 4y² = 12 be such that the length of latus rectum of H is equal to the length of latus rectum of E. If e_H and e_E are the eccentricities of H and E respectively, then the value of 12(e_H²+e_E²) is equal to ______.     (JEE Main 2022)

Ans. 42

∴ Length of latus rectum = 2/a

Length of latus rectum = 6/2 = 3

Q.52. Let the eccentricity of the hyperbola  be (5/4). If the equation of the normal at the point on the hyperbola is 8√5x + βy = λ, then λ − β is equal to ____.     (JEE Main 2022)

Ans. 85


Also  lies on the given hyperbola
So,
Equation of normal


⇒ 8√5x + 15y = 100
So, β = 15 and λ = 100
Gives λ − β = 85

Q.53. Let a line L₁ be tangent to the hyperbola  and let L₂ be the line passing through the origin and perpendicular to L₁. If the locus of the point of intersection of L₁ and L₂ is (x²+y²)² = αx² + βy², then α + β is equal to ____.      (JEE Main 2022)

Ans. 12
Equation of L₁ is

Equation of line L₂ is

∵ Required point of intersection of L₁ and L₂ is (x₁, y₁) then


From equations (iii) and (iv)

∴ Required locus of (x₁, y₁) is
(x² + y²)² = 16x² − 4y²
∴ α = 16, β = −4
∴ α + β = 12

Q.54. Let the common tangents to the curves 4(x² + y²) = 9 and y² = 4x intersect at the point Q. Let an ellipse, centered at the origin O, has lengths of semi-minor and semi-major axes equal to OQ and 6, respectively. If e and l respectively denote the eccentricity and the length of the latus rectum of this ellipse, then l/e² is equal to _____.     (JEE Main 2022)

Ans. 4
Let y = mx + c is the common tangent
So
So equation of common tangents will be  which intersects at Q(−3, 0)
Major axis and minor axis of ellipse are 12 and 6.
So eccentricity
 and length of latus rectum = 2b²/a = 3
Hence,

Q.55. A circle of radius 2 unit passes through the vertex and the focus of the parabola y2 = 2x and touches the parabola y = (x − (1/4))² + α, where α > 0. Then (4α − 8)² is equal to ________.      (JEE Main 2022)

Ans. 63

Let the equation of circle be
x(x − (1/2)) + y² + λy = 0
⇒ x² + y² − (1/2)x + λy = 0
Radius =


∵ This circle and parabola y − α = (x − (1/4))² touch each other, so



⇒ (4α - 8)² = 63

Q.56. Let  be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is 4(2√2 + √14). If the eccentricity H is √11/2, then the value of a² + b² is equal to ___.      (JEE Main 2022)

Ans. 88




By (3) and (4)


⇒ a = 4√2 ⇒a² = 32 and b² = 56
⇒ a² + b² = 32 + 56 = 88

Q.57. 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?      (JEE Advanced 2021)
(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'

Ans. a, b and 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, ∠QPQ′ = π/2 (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

Q.58. 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)

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 M₁M₂ = (1/2)AA' = 4

Q.59. Let θ be the acute angle between the tangents to the ellipse (x²/9) + (y²/1) = 1 and the circle x² + y² = 3 at their point of intersection in the first quadrant. Then tanθ is equal to :     (JEE Main 2021)
(a) 5/(2√3)
(b) 2/√3
(c) 4/√3
(d) 2

Ans. b
The point of intersection of the curves  and x² + y² = 3 in the first quadrant is
Now slope of tangent to the ellipse  is

And slope of tangent to the circle at
So, if angle between both curves is θ then

= 2/√3 Option (b)

Q.60. Consider the parabola with vertex ((1/2), (3/4)) and the directrix y = 1/2. Let P be the point where the parabola meets the line x = −(1/2). If the normal to the parabola at P intersects the parabola again at the point Q, then (PQ)² is equal to :      (JEE Main 2021)
(a) 75/8
(b) 125/16
(c) 25/2
(d) 15/2

Ans. b


For x = -(1/2)

Now, y′ = 2(x − (12))
At x = -(1/2)
⇒ m_T = -2, m_N = (1/2)
Equation of normal is

y = (x/2) + 2
Now put y in equation (1)

⇒ x = 2 & -(1/2)
⇒ Q(2, 3)
Now, (PQ)² = 125/16 Option (b)

Q.61. An angle of intersection of the curves,  and x² + y² = ab, a > b, is :     (JEE Main 2021)
(a)
(b)
(c)
(d)

Ans. c



∴ 2x₁ + 2y₁y′ = 0

Here (x₁y₁) is point of intersection of both curves

Q.62. The locus of mid-points of the line segments joining (−3, −5) and the points on the ellipse       (JEE Main 2021)
(a) 9x² + 4y² + 18x + 8y + 145 = 0
(b) 36x² + 16y² + 90x + 56y + 145 = 0
(c) 36x² + 16y² + 108x + 80y + 145 = 0
(d) 36x² + 16y² + 72x + 32y + 145 = 0

Ans. c
General point on
given B(−3, −5)



⇒ 36x² + 16y² + 108x + 80y + 145 = 0

Q.63. The line 12x cos⁡θ + 5y sin⁡θ = 60 is tangent to which of the following curves?      (JEE Main 2021)
(a) x² + y² = 169
(b) 144x² + 25y² = 3600
(c) 25x² + 12y² = 3600
(d) x² + y² = 60

Ans. b
12x cos⁡θ + 5y sin⁡θ = 60

is tangent to
144x² + 25y² = 3600

Q.64. The length of the latus rectum of a parabola, whose vertex and focus are on the positive x-axis at a distance R and S (> R) respectively from the origin, is :     (JEE Main 2021)
(a) 4(S + R)
(b) 2(S − R)
(c) 4(S − R)
(d) 2(S + R)

Ans. c

V → Vertex
F → focus
VF = S − R
So, latus rectum = 4(S − R)

Q.65. If two tangents drawn from a point P to the parabola y² = 16(x − 3) are at right angles, then the locus of point P is :     (JEE Main 2021)
(a) x + 3 = 0
(b) x + 1 = 0
(c) x + 2 = 0
(d) x + 4 = 0

Ans. b
Locus is directrix of parabola
x − 3 + 4 = 0 ⇒ x + 1 = 0.

Q.66. If x² + 9y² − 4x + 3 = 0, x, y ∈ R, then x and y respectively lie in the intervals :     (JEE Main 2021)
(a) [-(1/3), (1/3)] and [-(1/3), (1/3)]
(b) [-(1/3), (1/3)] and [1, 3]
(c) (1/3) and (1/3)
(d) (1/3) and [-(1/3), (1/3)]

Ans. d
x² + 9y² − 4x + 3 = 0
(x² − 4x) + (9y²) + 3 = 0
(x² − 4x + 4) + (9y²) + 3 − 4 = 0
(x − 2)² + (3y)² = 1
((x−2)²/(1)²) + (y²(1/3)²) = 1 (equation of an ellipse).
As it is equation of an ellipse, x & y can vary inside the ellipse.
So, x − 2 ∈ [−1,1] and y ∈ [-(1/3), (1/3)]
x ∈ [1, 3] y ∈ [-(1/3), (1/3)].

Q.67. A tangent and a normal are drawn at the point P(2, −4) on the parabola y2 = 8x, which meet the directrix of the parabola at the points A and B respectively. If Q(a, b) is a point such that AQBP is a square, then 2a + b is equal to :     (JEE Main 2021)
(a) -16
(b) -18
(c)-12
(d) -20

Ans. a

Given, parabola
y² = 8x ...... (i)
Equation of tangent at P(2,−4) is
−4y = 4(x+2)
or, x + y + 2 = 0 ..... (ii)
and Equation of normal to the parabola is
x − y + C = 0
∴ Normal passes through (2, −4)
∴ C = −6
Normal : x − y = 6 ..... (iii)
Equation of directrix of parabola
x = −2 ..... (iv)
Point of intersection of tangent and normal with directrix are x = −2 at A(−2, 0) and B(−2, −8) respectively.
Q(a, b) and P(2, −4) are given and AQBP is a square.
Mid-point of AB = Mid-point of PQ
⇒ (−2, −4) = ((a + 2)/2), (b − 4)/2)) ⇒ a = −6, b = −4
⇒ 2a + b = −16

Q.68. The locus of the mid points of the chords of the hyperbola x² − y² = 4, which touch the parabola y² = 8x, is :     (JEE Main 2021)
(a) y³(x − 2) = x²
(b) x³(x − 2) = y²
(c) y²(x − 2) = x³
(d) x²(x − 2) = y³

Ans. c
T = S₁
xh − yk = h₂ − k₂

this touches y² = 8x then c = a/m

2y² = x(y² − x²)
y²(x − 2) = x³

Q.69. The point P(−2√6, √3) lies on the hyperbola (x²/a²) − (y²/b²) = 1 having eccentricity √5/2. If the tangent and normal at P to the hyperbola intersect its conjugate axis at the point Q and R respectively, then QR is equal to :    (JEE Main 2021)
(a) 4√3
(b) 6
(c) 6√3
(d) 3√6

Ans. c
P(−2√6, √3) lies on hyperbola



⇒ a = √12

 
Tangent at P :

Slope of T = -(1/√2)
Normal at P :

⇒ R = (0, 5√3)
QR = 6√3

Q.70. On the ellipse  let P be a point in the second quadrant such that the tangent at P to the ellipse is perpendicular to the line x + 2y = 0. Let S and S' be the foci of the ellipse and e be its eccentricity. If A is the area of the triangle SPS' then, the value of (5 − e²). A is :    (JEE Main 2021)
(a) 6
(b) 12
(c) 14
(d) 24

Ans. a

Equation of tangent : y = 2x + 6 at P
∴ P(−8/3, 2/3)
e = 1/√2
S & S' = (−2, 0) & (2, 0)
Area of ΔSPS' = (1/2) × 4 × (2/3)
A = 4/3
∴ (5 − e²)A = (5 −(1/2))(4/3) = 6

Q.71. Let P and Q be two distinct points on a circle which has center at C(2, 3) and which passes through origin O. If OC is perpendicular to both the line segments CP and CQ, then the set {P, Q} is equal to :     (JEE Main 2021)
(a) {(4, 0), (0, 6)}
(b) {(2 + 2√2, 3 − √5),(2 − 2√2, 3 + √5)}
(c) {(2 + 2√2, 3 + √5),(2 − 2√2, 3 − √5)}
(d) {(−1, 5), (5, 1)}

Ans. d

tanθ = -(2/3)
Using symmetric from of line
P, Q:(2 ± √13cos⁡θ, 3 ±√13sin⁡θ)

(−1, 5) & (5, 1)

Q.72. A ray of light through (2, 1) is reflected at a point P on the y-axis and then passes through the point (5, 3). If this reflected ray is the directrix of an ellipse with eccentricity 1/3 and the distance of the nearer focus from this directrix is 8/√53, then the equation of the other directrix can be :    (JEE Main 2021)
(a) 11x + 7y + 8 = 0 or 11x + 7y − 15 = 0
(b) 11x − 7y − 8 = 0 or 11x + 7y + 15 = 0
(c) 2x − 7y + 29 = 0 or 2x − 7y − 7 = 0
(d) 2x − 7y − 39 = 0 or 2x − 7y − 7 = 0

Ans. c

Equation of reflected Ray
y−1 = (2/7)(x+2)
7x−7 = 2x+4
2x−7y+11 = 0
Let the equation of other directrix is
2x−7y+λ = 0
Distance of directrix from focus
(a/e) −e = 8/√53

Distance from other focus (a/e) + ae

Distance between two directrix = 2a/e

λ − 11 = 18 or − 18
λ = 29 or −7
2x − 7y − 7 = 0 or 2x − 7y + 29 = 0

Q.73. If a tangent to the ellipse x² + 4y² = 4 meets the tangents at the extremities of it major axis at B and C, then the circle with BC as diameter passes through the point :     (JEE Main 2021)
(a) (√3,0)
(b) (√2,0)
(c) (1, 1)
(d) (−1, 1)

Ans. a


Equation of tangent i (cosθ)x + 2sinθy = 2



Equation of circle is


so, (√3, 0) satisfying option (1)

Q.74. Let an ellipse  passes through  and has eccentricity 1/√3.  If a circle, centered at focus F(α, 0), α > 0, of E and radius 2√3, intersects E at two points P and Q, then PQ² is equal to :     (JEE Main 2021)
(a) 8/3
(b) 4/3
(c) 16/3
(d) 3

Ans. c

⇒ a² = 3b² = 3

Its focus is (1, 0)
Now, eqn of circle is
(x−1)² + y² = (4/3) ..... (ii)
Solving (i) and (ii) we get

Q.75. Let a parabola b be such that its vertex and focus lie on the positive x-axis at a distance 2 and 4 units from the origin, respectively. If tangents are drawn from O(0, 0) to the parabola P which meet P at S and R, then the area (in sq. units) of ΔSOR is equal to :    (JEE Main 2021)
(a) 16√2
(b) 16
(c) 32
(d) 8√2

Ans. b

Clearly RS is latus-rectum
∵ VF = 2 = a
∴ RS = 4a = 8
Now OF = 2a = 4
⇒ Area of triangle ORS = 16

Q.76. The locus of the centroid of the triangle formed by any point P on the hyperbola 16x² − 9y² +3 2x + 36y − 164 = 0, and its foci is :    (JEE Main 2021)
(a) 16x² − 9y² + 32x + 36y − 36 = 0
(b) 9x² − 16y² + 36x + 32y − 144 = 0
(c) 16x² − 9y² + 32x + 36y − 144 = 0
(d) 9x² − 16y² + 36x + 32y − 36 = 0

Ans. a
Given hyperbola is
16(x+1)² − 9(y−2)² = 164+16−36 = 144

Eccentricity,
⇒ foci are (4, 2) and (−6, 2)

Let the centroid be (h, k) & A(α, β) be point on hyperbola.

⇒ α = 3h + 2, β = 3k − 4
(α, β) lies on hyperbola so
16(3h + 2 + 1)² − 9(3k − 4 − 2)² = 144
⇒ 144(h + 1)² − 81(k − 2)² = 144
⇒ 16(h² + 2h + 1)− 9(k² − 4k + 4) = 16
⇒ 16x² − 9y² + 32x + 36y − 36 = 0

Q.77. Let  Let E₂ be another ellipse such that it touches the end points of major axis of E₁ and the foci of E₂ are the end points of minor axis of E₁. If E₁ and E₂ have same eccentricities, then its value is :    (JEE Main 2021)
(a)
(b)
(c)
(d)

Ans. a





Also b = ce
⇒ c = b/e


⇒ e² + e - 1 = 0
 

Q.78. Let a line L : 2x + y = k, k > 0 be a tangent to the hyperbola x² − y² = 3. If L is also a tangent to the parabola y² = αx, then α is equal to :    (JEE Main 2021)
(a) 12
(b) -12
(c)24
(d) -24

Ans. d
Tangent to hyperbola of
Slope m = −2 (given)


⇒ y + 2x = ± 3 ⇒ 2x + y = 3 (k > 0)
For parabola y² = αx
y = mx + (α/4m)
⇒ y = −2x + (α/−8)
⇒ (α/−8) =3
⇒ α = −24

Q.79. Let P be a variable point on the parabola y = 4x² + 1. Then, the locus of the mid-point of the point P and the foot of the perpendicular drawn from the point P to the line y = x is :      (JEE Main 2021)
(a) (3x−y)² + (x−3y) + 2 = 0

(b) 2(3x−y)² + (x−3y) + 2 = 0

(c) (3x−y)² + 2(x−3y) + 2 = 0

(d) 2(x−3y)² + (3x−y) + 2 = 0

Ans. b
Given, parabola y = 4x² + 1

Let R(a, b) be mid-point of line joining point P and Q where PQ is perpendicular to line y = x.
Let coordinates of P be P(x, y), Q(q, q) and R(a, b) then,

Now, slope of line y = x is m₁ = 1
Slope of line PQ be

∵ Line y = x and PQ are perpendicular to each other,
m₁ . m₂ = −1






Put (x, y) in equation of parabola as P(x, y) is variable point on parabola

Replace (a, b) as (x, y) ⇒ (3y − x) = 2(3x − y)² + 2
or 2(3x−y)² + (x−3y) + 2 = 0

Q.80. Let the tangent to the parabola S : y² = 2x at the point P(2, 2) meet the x-axis at Q and normal at it meet the parabola S at the point R. Then the area (in sq. units) of the triangle PQR is equal to :       (JEE Main 2021)
(a) 25/2
(b) 35/2
(c) 15/2
(d) 25

Ans. a

Tangent at P : y(2) = 2(1/2) (x + 2)
⇒ 2y = x + 2
∴ Q = (−2, 0)
Normal at P : y − 2 = -
⇒ y − 2 = − 2(x − 2)
⇒ y = 6 − 2x
∴ Solving with y² = 2x ⇒ R((9/2) − 3)

= (25/2) sq. units.

Q.81. Let a tangent be drawn to the ellipse (x²/27) + y² = 1 at (3√3cos⁡θ, sin⁡θ) where 0 ∈ (0, (π/2)). Then the value of θ such that the sum of intercepts on axes made by this tangent is minimum is equal to :     (JEE Main 2021)
(a) π/6
(b) π/3
(c) π/8
(d) π/4

Ans. a
Tangent
x-intercept = 3√3 secθ
y-intercept = cosecθ
sum = 3√3 secθ + cosecθ = f(θ) θ ∈ (0, (π/2))
⇒ f'(θ) = 3√3 secθ tanθ − cosecθ cotθ = 0




also f'(θ) changes sign − to + hence minimum.

Q.82. Consider a hyperbola H : x² − 2y² = 4. Let the tangent at a point P(4, √6) meet the x-axis at Q and latus rectum at R(x₁, y₁), x₁ > 0. If F is a focus of H which is nearer to the point P, then the area of ΔQFR is equal to :      (JEE Main 2021)
(a) √6 - 1
(b) (7√6) - 2
(c) (4√6) - 1
(d) 4√6

Ans. b

Given,
x² - 2y² = 4

Here, a = 2, b = √2

So, Focus (F) = (± a e, 0) = (±√6, 0)
Now, equation of tangent at P(4, √6) is
xx₁ − 2yy₁ = 4
⇒ x.4 − 2y.√6 = 4
⇒ 4x − 2√6y = 4
⇒ 2x − √6y = 2 ....... (i)
Putting y = 0 in Eq. (i), we get x-intercept of tangent i.e. x = 1
∴ Q ≡ (1, 0)
Hence, equation of corresponding latus rectum is x = √6

∴ Area of ΔQFR = (1/2) × (QF) × (RF)

Q.83. Let L be a tangent line to the parabola y² = 4x − 20 at (6, 2). If L is also a tangent to the ellipse  then the value of b is equal to:      (JEE Main 2021)
(a) 20
(b) 14
(c) 16
(d) 11

Ans. b
Parabola y² = 4x − 20
Tangent at P(6, 2) will be
2y = 4((x+6)/2)−20
2y = 2x + 12 − 20
2y = 2x − 8
y = x − 4
x − y − 4 = 0 ....... (1)
This is also tangent to ellipse (x²/2) + (y²/b) = 1
Apply c² = a²m² + b²
(−4)² = (2)(1) + b
b = 14

Q.84. Let C be the locus of the mirror image of a point on the parabola y² = 4x with respect to the line y = x. Then the equation of tangent to C at P(2, 1) is :     (JEE Main 2021)
(a) x − y = 1
(b) 2x + y = 5
(c) x + 3y = 5
(d) x + 2y = 4

Ans. a

Image of y² = 4x w.r.t. y = x is x² = 4y
tangent from (2, 1)
xx₁ = 2(y + y₁)
2x = 2(y + 1)
x = y + 1

Q.85. If the points of intersections of the ellipse  and the circle x² + y² = 4b, b > 4 lie on the curve y² = 3x², then b is equal to :     (JEE Main 2021)
(a) 12
(b) 10
(c) 6
(d) 5

Ans. a

x² + y² = 4b .... (2)
y² = 3x² .... (3)
From eq (2) and (3)
x² = b and y² = 3b
From equation (1)

⇒ b² + 48 = 16b
⇒ b = 12

Q.86. The locus of the midpoints of the chord of the circle, x² + y² = 25 which is tangent to the hyperbola,  is :     (JEE Main 2021)
(a) (x² + y²)² − 9x² + 16y² = 0
(b) (x² + y²)² − 9x² + 144y² = 0
(c) (x² + y²)² − 16x² + 9y² = 0
(d) (x² + y²)² − 9x² − 16y² = 0

Ans. a
tangent of hyperbola

which is a chord of circle with mid-point (h, k)
so equation of chord T = S₁
hx + ky = h² + k²

by (i) and (ii)


∴ Locus : 9x² − 16y² = (x² + y²)²

Q.87. If the three normals drawn to the parabola, y² = 2x pass through the point (a, 0) a ≠ 0, then 'a' must be greater than :     (JEE Main 2021)
(a) 1/2
(b) 1
(c) -1
(d) -(1/2)

Ans. b
Let the equation of the normal is
y = mx − 2am − am³
here 4a = 2 ⇒ a = (1/2)
y = mx − m − (1/2)m³
It passing through A(a, 0) then
0 = am − m − (1/2)m³
m = 0, a − 1 − (1/2)m² = 0
m² = 2(a − 1) > 0
⇒ a > 1

Q.88. A hyperbola passes through the foci of the ellipse (x²/25) + (y²/16) = 1 and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is :     (JEE Main 2021)
(a)
(b)
(c)
(d) x² - y² = 9

Ans. b

Foci = (±3, 0)
Let equation of hyperbola be
Passes through (±3, 0)
A² = 9, A = 3, e² = (5/3)


Equation of the hyperbola

Q.89. The shortest distance between the line x − y = 1 and the curve x² = 2y is :     (JEE Main 2021)
(a) 0
(b) 1/2√2
(c) 1/√2
(d) 1/2

Ans. b
Shortest distance must be along common normal

Let a point on curve has x coordinate = h
Then y coordinate :
h² = 2y
⇒ y = (h²/2)
So point is = (h, (h²/2))
m1 (slope of line x − y = 1) = 1
⇒ slope of perpendicular line = −1
Slope of the perpendicular line on the curve x² = 2y,
m₂ = 2x/2 = x ⇒ m₂ = h
∴ Slope of normal = −1h
−(1/h) = −1 ⇒ h = 1
So point is (1, (1/2))

Q.90. A tangent is drawn to the parabola y² = 6x which is perpendicular to the line 2x + y = 1. Which of the following points does NOT lie on it?     (JEE Main 2021)
(a) (0, 3)
(b) (-6, 0)
(c) (4, 5)
(d) (5, 4)

Ans. d
Equation of tangent : y = mx + (3/2m)
m_T = 1/2 (∵ perpendicular to line 2x + y = 1)
∴ tangent is : y = (x/2) + 3
⇒ x - 2y + 6 = 0

Q.91. If P is a point on the parabola y = x² + 4 which is closest to the straight line y = 4x − 1, then the co-ordinates of P are :      (JEE Main 2021)
(a) (−2, 8)
(b) (2, 8)
(c) (1, 5)
(d) (3, 13)

Ans. b
Given, curve y = x² + 4
and, line y = 4x − 1
Here, y = x² + 4

∴ (dy/dx) = 2x ..... (i)
and y = 4x − 1
(dy/dx) = 4 ..... (ii)
Let the required point be P(x1, y1).
∴ dy/dx|_P = 2x₁ ..... (iii)
∵ Slopes will be equal.
∴ 2x₁ = 4 [from Eqs. (ii) and (iii)]
⇒ x₁ = 4/2 =2
Now, the given point P(x₁, y₁) lies on curve y = x² + 4,
∴ y₁ = x₁² + 4
⇒ y₁ = 2² + 4 = 8
Hence, required coordinate of P = (2, 8)

Q.92. The locus of the mid-point of the line segment joining the focus of the parabola y² = 4ax to a moving point of the parabola, is another parabola whose directrix is :     (JEE Main 2021)
(a) x = 0
(b) x = -(a/2)
(c) x = a
(d) x = (a/2)

Ans. a
Given, equation of parabola ⇒ y² = 4ax
Focus = S(a, 0)
Let any point on the parabola be P(at², 2at).

and let the mid-point of PS be M(h, k).

Now,

∴ Locus of (h, k) is y² = a(2x−a)

∴ The directrix of this parabola is

Q.93. A tangent line L is drawn at the point (2, −4) on the parabola y² = 8x. If the line L is also tangent to the circle x² + y² = a, then 'a' is equal to ______.    (JEE Main 2021)

Ans. 2
tangent of y² = 8x is y = mx + (2/m)
P(2, −4) ⇒ −4 = 2m + (2/m)
⇒ m + (1/m) = −2 ⇒ m = −1
∴ tangent is y = −x −2
⇒ x + y + 2 = 0 ...... (1)
(1) is also tangent to x2 + y2 = a
So,
⇒ a = 2

Q.94. Let A (secθ, 2tanθ) and B (secϕ, 2tanϕ), where θ + ϕ = π/2, be two points on the hyperbola 2x² − y² = 2. If (α, β) is the point of the intersection of the normals to the hyperbola at A and B, then (2β)² is equal to ______.     (2021)    (JEE Main 2021)

Ans. Since, point A (secθ, 2tanθ) lies on the hyperbola 2x² − y² = 2
Therefore, 2sec²θ − 4tan²θ = 2
⇒ 2 + 2tan²θ − 4tan²θ = 2
⇒ tanθ = 0 ⇒ θ = 0
Similarly, for point B, we will get ϕ = 0.
but according to question θ + ϕ = (π/2) which is not possible.
Hence, it must be a 'BONUS'.

Q.95. If the minimum area of the triangle formed by a tangent to the ellipse  and the co-ordinate axis is kab, then k is equal to _______.     (JEE Main 2021)

Ans. 2
Tangent


So, area

⇒ k = 2

Q.96. Let E be an ellipse whose axes are parallel to the co-ordinates axes, having its center at (3, −4), one focus at (4, −4) and one vertex at (5, −4). If mx − y = 4, m > 0 is a tangent to the ellipse E, then the value of 5m² is equal to ______.     (JEE Main 2021)

Ans. 3
Given C(3, −4), S(4, −4)

and A(5, −4)
Hence, a = 2 & ae = 1
⇒ e = (1/2)
⇒ b² = 3
So,
Intersecting with given tangent.

Now, D = 0 (as it is tngent)
So, 5m² = 3.

Q.97. If the point on the curve y² = 6x, nearest to the point (3, (3/2)) is (α, β), then 2(α + β) is equal to ________.       (JEE Main 2021)

Ans. 9
Let, P ≡ ((3/2)t², 3t) is on the curve.
Normal at point P
tx + y  = 3t + (3/2)t³
Passes through (3, (3/2))

⇒ t³ = 1 ⇒ t = 1
P ≡ ((3/2), 3) = (α, β)
2(α+β) = 2((3/2)+3) = 9

Q.98. Let y = mx + c, m > 0 be the focal chord of y² = − 64x, which is tangent to (x + 10)² + y² = 4. Then, the value of 4√2 (m + c) is equal to ________.      (JEE Main 2021)

Ans. 34
y² = −64x
focus : (−16, 0)
y = mx + c is focal chord
⇒ c = 16 m .....(1)
y = mx + c is tangent to (x + 10)² + y² = 4




⇒ 9m² = 1 + m²

Q.99. Let L be a common tangent line to the curves 4x² + 9y² = 36 and (2x)² + (2y)² = 31. Then the square of the slope of the line L is _____.     (JEE Main 2021)

Ans. 3
Tangent to the curve

and equation of tangent to the curve x² + y² = (31/4) is

for common tangent

⇒ m² =3

Q.100. A line is a common tangent to the circle (x − 3)² + y² = 9 and the parabola y² = 4x. If the two points of contact (a, b) and (c, d) are distinct and lie in the first quadrant, then 2(a + c) is equal to _________.       (JEE Main 2021)

Ans. 9
Circle : (x − 3)² + y² = 9
Parabola : y² = 4x
Let tangent y = mx + (a/m)
y = mx + (1/m)
m²x − my + 1 = 0
the above line is also tangent to circle
(x − 3)² + y² = 9
∴ ⊥ from (3, 0) = 3

(3m² + 1)² = 9(m² + m⁴)
6m² + 1 + 9m⁴ = 9m² + 9m⁴
3m² = 1
m = ±1/√3
∴ tangent is

(it will be used)
or

(rejected)


For parabola

for circle
&
(x−3)² + y² = 9
Solving,



4x² − 12x + 9 = 0
4x² − 6x − 6x + 9 = 0
2x(2x − 3) − 3(2x − 3) = 0
(2x − 3)(2x − 3) = 0
x = 3/2

Q.101. The locus of the point of intersection of the lines (√3)kx + ky − 4√3 = 0 and √3x − y −4(√3)k = 0 is a conic, whose eccentricity is _________.       (JEE Main 2021)

Ans. 2


Adding equation (1) & (2)


Substracting equation (1) & (2)




⇒ e = 2.