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Page 1
JEE Main Previous Year Questions
(2025): Ellipse
Q1: If ?? and ?? '
are the foci of the ellipse
?? ?? ????
+
?? ?? ?? = ?? and P be a point on the ellipse,
then ?????? ( ???? . ?? '
?? )+ ?????? ( ???? . ?? '
?? ) is equal to :
A. 3( 1 + v 2)
B. 3( 6 + v 2)
C. 9
D. 27
Ans: D
Solution:
PS + PS
'
= 2 × 3v 2
b
2
= a
2
( 1 - e
2
) ? 9 = 18( 1 - e
2
)
? e =
1
v 2
Directrix x =
a
e
=
3v 2
1
v 2
= 6
???? · ?? ?? '
= |
1
v 2
( 3v 2cos ?? - 6)
1
v 2
( 3v 2cos ?? + 6) |
=
1
2
|18cos
2
?? - 36|
( PS · PS
'
)
max
= 18; ( PS · PS)
min
= 9
sum = 27
Page 2
JEE Main Previous Year Questions
(2025): Ellipse
Q1: If ?? and ?? '
are the foci of the ellipse
?? ?? ????
+
?? ?? ?? = ?? and P be a point on the ellipse,
then ?????? ( ???? . ?? '
?? )+ ?????? ( ???? . ?? '
?? ) is equal to :
A. 3( 1 + v 2)
B. 3( 6 + v 2)
C. 9
D. 27
Ans: D
Solution:
PS + PS
'
= 2 × 3v 2
b
2
= a
2
( 1 - e
2
) ? 9 = 18( 1 - e
2
)
? e =
1
v 2
Directrix x =
a
e
=
3v 2
1
v 2
= 6
???? · ?? ?? '
= |
1
v 2
( 3v 2cos ?? - 6)
1
v 2
( 3v 2cos ?? + 6) |
=
1
2
|18cos
2
?? - 36|
( PS · PS
'
)
max
= 18; ( PS · PS)
min
= 9
sum = 27
Q2: 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
v 17
B.
v 3
16
C.
3
v 19
D.
v 5
7
Ans: A
Solution:
2?? =
1
4
( 2???? )
?? ?? =
?? 4
?? =
v
1 -
?? 2
?? 2
?? =
v
1 -
?? 2
16
?? 2
(1 +
1
16
) = 1
?? =
4
v 17
Q3: A line passing through the point ?? ( v?? , v?? ) intersects the ellipse
?? ?? ????
+
?? ?? ????
= ?? at ??
and ?? such that ( ???? )· ( ???? ) is maximum. Then ?? ( ?? ?? ?? + ?? ?? ?? ) is equal to :
A. 218
B. 377
C. 290
D. 338
Ans: D
Solution: Given ellipse is x
2
/ 36 + y
2
/ 25 = 1
Page 3
JEE Main Previous Year Questions
(2025): Ellipse
Q1: If ?? and ?? '
are the foci of the ellipse
?? ?? ????
+
?? ?? ?? = ?? and P be a point on the ellipse,
then ?????? ( ???? . ?? '
?? )+ ?????? ( ???? . ?? '
?? ) is equal to :
A. 3( 1 + v 2)
B. 3( 6 + v 2)
C. 9
D. 27
Ans: D
Solution:
PS + PS
'
= 2 × 3v 2
b
2
= a
2
( 1 - e
2
) ? 9 = 18( 1 - e
2
)
? e =
1
v 2
Directrix x =
a
e
=
3v 2
1
v 2
= 6
???? · ?? ?? '
= |
1
v 2
( 3v 2cos ?? - 6)
1
v 2
( 3v 2cos ?? + 6) |
=
1
2
|18cos
2
?? - 36|
( PS · PS
'
)
max
= 18; ( PS · PS)
min
= 9
sum = 27
Q2: 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
v 17
B.
v 3
16
C.
3
v 19
D.
v 5
7
Ans: A
Solution:
2?? =
1
4
( 2???? )
?? ?? =
?? 4
?? =
v
1 -
?? 2
?? 2
?? =
v
1 -
?? 2
16
?? 2
(1 +
1
16
) = 1
?? =
4
v 17
Q3: A line passing through the point ?? ( v?? , v?? ) intersects the ellipse
?? ?? ????
+
?? ?? ????
= ?? at ??
and ?? such that ( ???? )· ( ???? ) is maximum. Then ?? ( ?? ?? ?? + ?? ?? ?? ) is equal to :
A. 218
B. 377
C. 290
D. 338
Ans: D
Solution: Given ellipse is x
2
/ 36 + y
2
/ 25 = 1
Any point on line ???? can be assumed as
Q( v 5 + rcos ?? , v 5 + rsin ?? )
Putting this in equation of ellipse, we get
25( v 5 + ?? cos ?? )
2
+ 36( v 5 + ?? sin ?? )
2
= 900
Simplifying, we get
r
2
( 25cos
2
?? + 36sin
2
?? )+ 2v 5r( 25cos ?? + 36sin ?? )- 595 = 0
|r| = PA, PB
PA · PB =
595
25cos
2
?? + 36sin
2
?? =
595
25 + 11sin
2
?? = maximum, if sin
2
?? = 0
This means line ???? must be parallel to ?? -axis
? ?? ?? = ?? ?? = v 5
Putting ?? = v 5 in equation of ellipse, we get
?? 2
36
+
1
5
= 1 ? ?? 2
= 36 ·
4
5
Hence,
PA
2
+ PB
2
= (v 5 -
12
v 5
)
2
+ (v 5 +
12
v 5
)
2
= 2 (5 +
144
5
) =
338
5
5( PA
2
+ PB
2
) = 338
Q4: The length of the latus-rectum of the ellipse, whose foci are ( ?? , ?? ) and ( ?? , -?? )
and eccentricity is
?? ?? , is
Page 4
JEE Main Previous Year Questions
(2025): Ellipse
Q1: If ?? and ?? '
are the foci of the ellipse
?? ?? ????
+
?? ?? ?? = ?? and P be a point on the ellipse,
then ?????? ( ???? . ?? '
?? )+ ?????? ( ???? . ?? '
?? ) is equal to :
A. 3( 1 + v 2)
B. 3( 6 + v 2)
C. 9
D. 27
Ans: D
Solution:
PS + PS
'
= 2 × 3v 2
b
2
= a
2
( 1 - e
2
) ? 9 = 18( 1 - e
2
)
? e =
1
v 2
Directrix x =
a
e
=
3v 2
1
v 2
= 6
???? · ?? ?? '
= |
1
v 2
( 3v 2cos ?? - 6)
1
v 2
( 3v 2cos ?? + 6) |
=
1
2
|18cos
2
?? - 36|
( PS · PS
'
)
max
= 18; ( PS · PS)
min
= 9
sum = 27
Q2: 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
v 17
B.
v 3
16
C.
3
v 19
D.
v 5
7
Ans: A
Solution:
2?? =
1
4
( 2???? )
?? ?? =
?? 4
?? =
v
1 -
?? 2
?? 2
?? =
v
1 -
?? 2
16
?? 2
(1 +
1
16
) = 1
?? =
4
v 17
Q3: A line passing through the point ?? ( v?? , v?? ) intersects the ellipse
?? ?? ????
+
?? ?? ????
= ?? at ??
and ?? such that ( ???? )· ( ???? ) is maximum. Then ?? ( ?? ?? ?? + ?? ?? ?? ) is equal to :
A. 218
B. 377
C. 290
D. 338
Ans: D
Solution: Given ellipse is x
2
/ 36 + y
2
/ 25 = 1
Any point on line ???? can be assumed as
Q( v 5 + rcos ?? , v 5 + rsin ?? )
Putting this in equation of ellipse, we get
25( v 5 + ?? cos ?? )
2
+ 36( v 5 + ?? sin ?? )
2
= 900
Simplifying, we get
r
2
( 25cos
2
?? + 36sin
2
?? )+ 2v 5r( 25cos ?? + 36sin ?? )- 595 = 0
|r| = PA, PB
PA · PB =
595
25cos
2
?? + 36sin
2
?? =
595
25 + 11sin
2
?? = maximum, if sin
2
?? = 0
This means line ???? must be parallel to ?? -axis
? ?? ?? = ?? ?? = v 5
Putting ?? = v 5 in equation of ellipse, we get
?? 2
36
+
1
5
= 1 ? ?? 2
= 36 ·
4
5
Hence,
PA
2
+ PB
2
= (v 5 -
12
v 5
)
2
+ (v 5 +
12
v 5
)
2
= 2 (5 +
144
5
) =
338
5
5( PA
2
+ PB
2
) = 338
Q4: The length of the latus-rectum of the ellipse, whose foci are ( ?? , ?? ) and ( ?? , -?? )
and eccentricity is
?? ?? , is
A.
6
5
B.
50
3
C.
10
3
D.
18
5
Ans: D
Solution:
2be = 8
be = 4
?? (
4
5
) = 4 ? ?? = 5
? ?? 2
= ?? 2
- ?? 2
16 = 25 - ?? 2
? ?? = 3
L.R. =
2?? 2
?? =
18
5
Q5: The centre of a circle C is at the centre of the ellipse ?? :
?? ?? ?? ?? +
?? ?? ?? ?? = ?? , ?? > ?? . Let ??
pass through the foci ?? ?? and ?? ?? of ?? such that the circle ?? and the ellipse ?? intersect
at four points. Let P be one of these four points. If the area of the triangle ????
?? ?? ?? is 30
and the length of the major axis of E is 17 , then the distance between the foci of E is :
A. 26
B. 13
C. 12
D.
13
2
Ans: B
Solution:
Page 5
JEE Main Previous Year Questions
(2025): Ellipse
Q1: If ?? and ?? '
are the foci of the ellipse
?? ?? ????
+
?? ?? ?? = ?? and P be a point on the ellipse,
then ?????? ( ???? . ?? '
?? )+ ?????? ( ???? . ?? '
?? ) is equal to :
A. 3( 1 + v 2)
B. 3( 6 + v 2)
C. 9
D. 27
Ans: D
Solution:
PS + PS
'
= 2 × 3v 2
b
2
= a
2
( 1 - e
2
) ? 9 = 18( 1 - e
2
)
? e =
1
v 2
Directrix x =
a
e
=
3v 2
1
v 2
= 6
???? · ?? ?? '
= |
1
v 2
( 3v 2cos ?? - 6)
1
v 2
( 3v 2cos ?? + 6) |
=
1
2
|18cos
2
?? - 36|
( PS · PS
'
)
max
= 18; ( PS · PS)
min
= 9
sum = 27
Q2: 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
v 17
B.
v 3
16
C.
3
v 19
D.
v 5
7
Ans: A
Solution:
2?? =
1
4
( 2???? )
?? ?? =
?? 4
?? =
v
1 -
?? 2
?? 2
?? =
v
1 -
?? 2
16
?? 2
(1 +
1
16
) = 1
?? =
4
v 17
Q3: A line passing through the point ?? ( v?? , v?? ) intersects the ellipse
?? ?? ????
+
?? ?? ????
= ?? at ??
and ?? such that ( ???? )· ( ???? ) is maximum. Then ?? ( ?? ?? ?? + ?? ?? ?? ) is equal to :
A. 218
B. 377
C. 290
D. 338
Ans: D
Solution: Given ellipse is x
2
/ 36 + y
2
/ 25 = 1
Any point on line ???? can be assumed as
Q( v 5 + rcos ?? , v 5 + rsin ?? )
Putting this in equation of ellipse, we get
25( v 5 + ?? cos ?? )
2
+ 36( v 5 + ?? sin ?? )
2
= 900
Simplifying, we get
r
2
( 25cos
2
?? + 36sin
2
?? )+ 2v 5r( 25cos ?? + 36sin ?? )- 595 = 0
|r| = PA, PB
PA · PB =
595
25cos
2
?? + 36sin
2
?? =
595
25 + 11sin
2
?? = maximum, if sin
2
?? = 0
This means line ???? must be parallel to ?? -axis
? ?? ?? = ?? ?? = v 5
Putting ?? = v 5 in equation of ellipse, we get
?? 2
36
+
1
5
= 1 ? ?? 2
= 36 ·
4
5
Hence,
PA
2
+ PB
2
= (v 5 -
12
v 5
)
2
+ (v 5 +
12
v 5
)
2
= 2 (5 +
144
5
) =
338
5
5( PA
2
+ PB
2
) = 338
Q4: The length of the latus-rectum of the ellipse, whose foci are ( ?? , ?? ) and ( ?? , -?? )
and eccentricity is
?? ?? , is
A.
6
5
B.
50
3
C.
10
3
D.
18
5
Ans: D
Solution:
2be = 8
be = 4
?? (
4
5
) = 4 ? ?? = 5
? ?? 2
= ?? 2
- ?? 2
16 = 25 - ?? 2
? ?? = 3
L.R. =
2?? 2
?? =
18
5
Q5: The centre of a circle C is at the centre of the ellipse ?? :
?? ?? ?? ?? +
?? ?? ?? ?? = ?? , ?? > ?? . Let ??
pass through the foci ?? ?? and ?? ?? of ?? such that the circle ?? and the ellipse ?? intersect
at four points. Let P be one of these four points. If the area of the triangle ????
?? ?? ?? is 30
and the length of the major axis of E is 17 , then the distance between the foci of E is :
A. 26
B. 13
C. 12
D.
13
2
Ans: B
Solution:
1
2
PF
1
· PF
2
= 30
PF
1
+ PF
2
= 17
PF
1
= 12PF
2
= 5
F
1
F
2
= 13
option ( 2)
Q6: Let ?? = {( ?? , ?? ) ? ?? × ?? : |?? - ?? | = ?? and |?? - ?? | = ?? } and ?? = {( ?? , ?? ) ? ?? ×
?? : ???? ( ?? - ?? )
?? + ?? ( ?? - ?? )
?? = ?????? }
A. B ? A
B. A ? B = {( x, y) : -4 = x = 4, -1 = y = 11}
C. neither A ? B nor B ? A
D. ?? ? ??
Ans: A
Solution:
A: |?? - 1| = 4 and |?? - 5| = 6
? -4 = ?? - 1 = 4 ? -6 = ?? - 5 = 6
? -3 = ?? = 5 ? -1 = ?? = 11
B: 16( ?? - 2)
2
+ 9( ?? - 6)
2
= 144
B :
( ?? - 2)
2
9
+
( ?? - 6)
2
16
= 1
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FAQs on JEE Main Previous Year Questions (2025): Ellipse - Mathematics (Maths) for JEE Main & Advanced
| 1. What is the standard equation of an ellipse and how is it derived? |  |
Ans. The standard equation of an ellipse centered at the origin is given by (x²/a²) + (y²/b²) = 1, where 'a' is the semi-major axis and 'b' is the semi-minor axis. This equation is derived from the definition of an ellipse as the set of points where the sum of the distances from two fixed points (foci) is constant. When the ellipse is aligned with the coordinate axes, the lengths of the axes can be represented by 'a' and 'b', leading to the standard form of the equation.
| 2. How do you identify the foci of an ellipse? | |
Ans. The foci of an ellipse can be found using the formula c² = a² - b², where 'c' is the distance from the center to each focus, 'a' is the length of the semi-major axis, and 'b' is the length of the semi-minor axis. The foci are located at coordinates (c, 0) and (-c, 0) for horizontal ellipses and (0, c) and (0, -c) for vertical ellipses.
| 3. What is the difference between a major axis and a minor axis in an ellipse? | |
Ans. The major axis of an ellipse is the longest diameter that passes through both foci, while the minor axis is the shortest diameter that is perpendicular to the major axis at the center. The lengths of these axes are represented by '2a' for the major axis and '2b' for the minor axis, where 'a' and 'b' are the lengths of the semi-major and semi-minor axes, respectively.
| 4. How do you find the area of an ellipse? | |
Ans. The area of an ellipse can be calculated using the formula A = πab, where 'a' is the semi-major axis and 'b' is the semi-minor axis. This formula arises from the geometric properties of the ellipse and represents the total space enclosed within it.
| 5. What are the applications of ellipses in real life? | |
Ans. Ellipses have various applications in real life, including in the design of orbits for celestial bodies, where planets follow elliptical paths around the sun. They are also used in engineering, architecture, and acoustics, such as in the design of elliptical buildings that focus sound. Additionally, ellipses are utilized in optics, where lenses are shaped to follow elliptical curves for better light focus.
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Oct 11, 2025
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