Students can find the JEE Advanced Maths ellipse previous years’ questions and solutions on this page. These are solved by our top JEE subject experts and students will reap the benefits of finding only accurate answers. Revising previous questions will definitely help students to be familiar with the type of problems asked in the JEE Advanced exam. This will improve the confidence levels of the students. Solutions are also available in PDF format, for free. Students can easily download and learn the solutions offline.
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JEE Advanced Previous Years’ Questions with Solutions on Ellipse
Question 1: If the point P on the curve, 4x2 + 5y2 = 20 is farthest from the point Q(0, -4), then PQ2 is equal to
(a) 36
(b) 48
(c) 21
(d) 29
Solution:
Given curve is 4x2 + 5y2 = 20
=> (x2/5) + (y2/4) = 1 (ellipse)
=> a = √5, b = 2
Let a point on the ellipse be P(a cos θ, b sin θ).
=> P = (√5 cos θ, 2 sin θ)
Q = (0, -4)
Using distance formula
PQ2 = (√5 cos θ)2 + (2 sin θ + 4)2
Let PQ2 = Z
=> Z = 5 cos2 θ + 16 + 16 sin θ + 4 sin2θ
= cos2 θ + 4cos2 θ+ 4 sin2θ + 16 + 16 sin θ
= 4 + 16 + cos2 θ + 16 sin θ
= 20 + cos2 θ + 16 sin θ
For maxima, dZ/dθ = 0
=> -2 cos θ sin θ + 16 cos θ = 0
=> cos θ = 0 or sin θ = 8
=> θ = π/2
d2Z/dθ2 < 0
So Z has maxima at θ = π/2
=> Z = 20 + cos2 π/2+ 16 sin π/2
= 20 + 0 + 16
= 36
Hence option a is the answer.
Question 2: The length of minor axis (along y-axis) of an ellipse of the standard form is 4/√3. If this ellipse touches the line x + 6y = 8, then its eccentricity is:
(a) (1/2)(√5/3)
(b) (1/2)√(11/3)
(c) √(5/6)
(d) (1/3)√(11/3)
Solution:
Given length of minor axis = 4/√3
2b = 4/√3
b = 2/√3
Equation of tangent is y = mx ±√(a2m2+b2)
Comparing y = (-x/6)+(8/6) with y = mx ±√(a2m2+b2)
We get m = -1/6
a2m2 + b2 = 16/9
(a2/36) + (4/3) = 16/9
(a2/36) = (16/9) – (4/3)
a2 = 16
e = √(1 – b2/a2)
e = √(11/12)
= (1/2)√(11/3)
Hence option b is the answer.
Question 3: The eccentricity of an ellipse whose centre is at the origin is 1/2. If one of its directrices is x = −4, then the equation of the normal to it at (1, (3 / 2)) is:
(a) 4x – 2y = 1
(b) 4x + 2y = 7
(c) x + 2y = 4
(d) 2y − x = 2
Solution:
Given that x = -4
e = 1/2
(-a/e) = -4
a = 4e
a = 2
Now, b2 = a2 (1 – e2)
= 3
Equation to ellipse is (x2/4) + (y2/3) = 1
Equation of normal is (x-1)/(¼) = (y-3/2)/(3/2×3)
=> 4x – 2y = 1
Hence option a is the answer.
Question 4: If the tangents on the ellipse 4x2+y2 = 8 at the points (1, 2) and (a, b) are perpendicular to each other, then a2 is equal to
(a) 2/17
(b) 64/17
(c) 128/17
(d) 4/17
Solution:
Given the equation of the ellipse 4x2+y2 = 8
dy/dx = -4x/y
The tangent at (1, 2) and (a, b) are perpendicular
(-y/2)(-4a/b) = -1
b = -8a ..(i)
(a,b) is on the ellipse.
4a2 + b2 = 8 [from eq(i)]
4a2 + 64a2 = 8
a2 = 8/68
= 2/17
Hence option a is the answer.
Question 5: Which of the following points lies on the locus of the foot of perpendicular drawn upon any tangent to the ellipse (x2/4) + (y2/2) = 1 from any of its foci?
(a) (-1,√3)
(b) (-2, √3)
(c) (-1, √2)
(d) (1, 2)
Solution:
Given (x2/4) + (y2/2) = 1
=> a = 2, b = √2
Let foot of perpendicular is (h, k)
e = √(1 – b2/a2)
= √(1- 2/4)
= 1/√2
So focus (ae,0) = (√2,0)
Equation of tangent is y = mx+√(a2m2 + b2)
y = mx+√(4m2+2)
Passes through (h,k)
(k – mh)2 = 4m2 + 2 ..(i)
The line perpendicular to tangent will have slope -1/m.
y – 0 = -(1/m)(x-√2)
my = -x+√2
(h+mk)2 = 2 ..(ii)
Add equation (i) and (ii)
k2(1+m2) + h2(1+m2) = 4(1+m2)
h2 + k2 = 4
x2 + y2 = 4 (auxiliary circle)
(-1,√3) lies on the locus.
Hence option a is the answer.
Question 6: 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:
(a) x2/9 – y2/4 = 1
(b) x2/9 – y2/16 = 1
(c) x2 – y2 = 9
(d) x2/9 – y2/25 = 1
Solution:
Given equation of ellipse is x2/25 + y2/16 = 1
=> a = 5, b = 4
For ellipse, e1 = √(1 – b2/a2)
= √(1-16/25)
=3/5
Foci = (±3,0)
Let equation of hyperbola be x2/A2 – y2/B2 = 1, passes through (±3, 0)
=> A2 = 9
A = 3
e2 = 5/3
e22 = 1 + B2/A2
25/9 = 1 + B2/9
B2 = 16
x2/9 – y2/16 = 1
Hence option b is the answer.
Question 7: If e1 and e2 are the eccentricities of the ellipse, (x2/18) + (y2/4) = 1 and the hyperbola, (x2/9) – (y2/4) = 1 respectively and (e1, e2) is a point on the ellipse, 15x2 + 3y2 = k. Then k is equal to:
(a) 14
(b) 15
(c) 17
(d) 16
Solution:
Given (x2/18) + (y2/4) = 1
=> a2 = 18, b2 = 4
Eccentricity, e = √(1 – b2/a2)
e1 = √(1-(4/18))
= √7/3
(x2/9) – (y2/4) = 1
e2 = √(1 + (4/9))
= √13/3
Since (e1, e2) lies on the ellipse 15x2 + 3y2 = k
15×(7/9) + 3×(13/9) = k
k = 16
Hence option d is the answer.
Question 8: Let the line y = mx and the ellipse 2x2+ y2 = 1 intersect a point P in the first quadrant. If the normal to this ellipse at P meets the co-ordinate axes at (-1/3√2, 0) and (0, β), then β is equal to
(a) 2/√3
(b) 2/3
(c) 2√2/3
(d) √2/3
Solution:
Let P ≡ (x1, y1)
2x2 + y2 = 1 is given equation of ellipse.
=> 4x + 2yy’= 0
=> y’|(x1, y1) = -2x1/y1
The slope of normal at (x1, y1) is y1/2x1
Equation of normal at (x1, y1) is (y – y1) = (y1/2x1) (x – x1)
It passes through (-1/3√2, 0)
=> -y1 = (y1/2x1)((-1/3√2) – x1)
=> x1 = 1/3√2
=> y1 = 2√2/3
as P lies in the first quadrant
Since (0, β) lies on the normal of the ellipse at point P, hence we get
β = y1/2
= √2/3
Hence option d is the answer.
Question 9: If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12, then the length of its latus rectum is:
(a) √3
(b) 3√2
(c) 3/√2
(d) 2√3
Solution:
Given that the distance between the foci is 6
=> 2ae = 6
=> ae = 3 ..(i)
Given that distance between directrices is 12
=> 2a/e = 12
=> a = 6e ..(ii)
6e2 = 3
=> e2 = ½
=> e = 1/√2
So a = 6/√2
= 3√2
We know e2 = 1 – b2/a2
=> ½ = 1 – b2/18
=> ½ = b2/18
=> b2 = 9
Length of latus rectum = 2b2/a
= 2×9/3√2
= 3√2
Hence option b is the answer.
Question 10: If the line x – 2y = 12 is tangent to the ellipse x2/a2 + y2/b2 = 1 at the point (3, -9/2) then the length of the latus rectum of the ellipse is
(a) 9
(b) 12√2
(c) 5
(d) 8√3
Solution:
Given that x – 2y = 12 is tangent to the ellipse x2/a2 + y2/b2 = 1 at (3, -9/2)
Equation of tangent to x2/a2 + y2/b2 = 1 at (3, -9/2)
3x/a2 – 9y/2b2 = 1 ..(i)
Comparing (i) with x – 2y = 12
We get 3/a2 = 9/4b2 = 1/12
=> a2 = 36
So a = 6
b2 = 12(9/4)
= 27
The length of the latus rectum of the ellipse is 2b2/a
= 2(27/6)
= 9
Hence option a is the answer.
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