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+ 
= 1, and its corresponding point Q on the auxiliary circle meet on the line
The radius of the circle given by 2x2 + 2y2 – x = 0 is- a)1/4
- b)1
- c)2
- d)1/2
Correct answer is option 'A'. Can you explain this answer?
The radius of the circle given by 2x2 + 2y2 – x = 0 is
a)
1/4
b)
1
c)
2
d)
1/2
|
Geetika Shah answered |
2x² + 2y² - x = 0 .
==> 2 ( x² + y² - x/2 ) = 0 .
==> 2/2 ( x² + y² - x/2 ) = 0/2 .
==> x² - x/2 + y² = 0 .
==> ( x² - x/2 + (1/4)² ) + y² = (1/4)² .
==> ( x - 1/4 )² + ( y - 0 )² = (1/4)²
Centre (-¼, 0) radius(¼)
==> 2 ( x² + y² - x/2 ) = 0 .
==> 2/2 ( x² + y² - x/2 ) = 0/2 .
==> x² - x/2 + y² = 0 .
==> ( x² - x/2 + (1/4)² ) + y² = (1/4)² .
==> ( x - 1/4 )² + ( y - 0 )² = (1/4)²
Centre (-¼, 0) radius(¼)
The centre of the ellipse 
is:- a)(0, 1)
- b)(1, 1)
- c)(0, 0)
- d)(1, 0)
Correct answer is option 'B'. Can you explain this answer?
The centre of the ellipse is:
is:a)
(0, 1)
b)
(1, 1)
c)
(0, 0)
d)
(1, 0)
|
Preeti Iyer answered |
Centre of the ellipse is the intersection point of
x+y−1=0.........(1)
x−y=0............(2)
Substituting x from equation 2 in equation 1 two equations, we get,
2y=2, y=1
Replacing, we get x=1
⇒(1,1) is the centre
x+y−1=0.........(1)
x−y=0............(2)
Substituting x from equation 2 in equation 1 two equations, we get,
2y=2, y=1
Replacing, we get x=1
⇒(1,1) is the centre
The equation of the ellipse whose one focus is at (4, 0) and whose eccentricity is 4/5 is:- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The equation of the ellipse whose one focus is at (4, 0) and whose eccentricity is 4/5 is:
a)
b)
c)
d)
|
Trisha Vashisht answered |
focus lies on x axis
So, the equation of ellipse is x2/a2 + y2/b2 = 1
Co-ordinate of focus(+-ae, 0)
ae = 4
e = ⅘
a = 4/e => 4/(⅘)
a = 5
(a)2 = 25
b2 = a2(1-e2)
= 25(1-16/25)
b2 = 9
Required equation : x2/(5)2 + y2/(3)2 = 1
So, the equation of ellipse is x2/a2 + y2/b2 = 1
Co-ordinate of focus(+-ae, 0)
ae = 4
e = ⅘
a = 4/e => 4/(⅘)
a = 5
(a)2 = 25
b2 = a2(1-e2)
= 25(1-16/25)
b2 = 9
Required equation : x2/(5)2 + y2/(3)2 = 1
If the curve x2 3y2 = 9 subtends an obtuse angle at the point (2α, α), then a possible value of α2 is
- a) 1
- b) 2
- c) 3
- d) 4
Correct answer is option 'A'. Can you explain this answer?
If the curve x2 3y2 = 9 subtends an obtuse angle at the point (2α, α), then a possible value of α2 is
a)
1b)
2c)
3d)
4
|
Ishan Choudhury answered |
The given curve is 
whose director circle is x2 + y2 = 12.
For the required condition (2α, α) should lie inside the circle and outside the ellipse i.e.,
whose director circle is x2 + y2 = 12.For the required condition (2α, α) should lie inside the circle and outside the ellipse i.e.,
If distance between the directrices be thrice the distance between the foci, then eccentricity of ellipse is- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
If distance between the directrices be thrice the distance between the foci, then eccentricity of ellipse is
a)
b)
c)
d)
|
Neha Sharma answered |
Distance between the directrices is 2a/e
Distance between the foci is 2ae
Given: 2a/e = 3∗2ae
Or,e2 = 1/3
∴ e=1/√3
Distance between the foci is 2ae
Given: 2a/e = 3∗2ae
Or,e2 = 1/3
∴ e=1/√3
The eccentricity of an ellipse whose latus rectum is equal to distance between foci is:- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
The eccentricity of an ellipse whose latus rectum is equal to distance between foci is:
a)
b)
c)
d)
|
Suresh Reddy answered |
Distance between the foci of an ellipse = length of latus rectum
i.e. (2b2)/a=2ae
e=b2/a2
But e=[1−b2/a2]1/2
Then e=(1−e)1/2
Squaring both sides, we get
e2 +e−1=0
e=−1 ± (1 + 4)1/2]/2
(∵ Eccentricity cannot be negative)
e=[(5)1/2 − 1]/2
i.e. (2b2)/a=2ae
e=b2/a2
But e=[1−b2/a2]1/2
Then e=(1−e)1/2
Squaring both sides, we get
e2 +e−1=0
e=−1 ± (1 + 4)1/2]/2
(∵ Eccentricity cannot be negative)
e=[(5)1/2 − 1]/2
From point P (8, 27), tangent PQ and PR are drawn to the ellipse 
Then the angle subtended by QR at origin is - a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
From point P (8, 27), tangent PQ and PR are drawn to the ellipse Then the angle subtended by QR at origin is
Then the angle subtended by QR at origin is a)
b)
c)
d)
|
Om Desai answered |
Equation of QR is T = 0 (chord of contact)
⇒ 2x + 3y = 1 .....(i)
Now, equation of the pair of lines passing through origin and points Q, R is given by
(making equation of ellipse homogeneous using Eq (i)
∴135x2 + 432xy + 320y2 = 0
⇒ 2x + 3y = 1 .....(i)
Now, equation of the pair of lines passing through origin and points Q, R is given by
(making equation of ellipse homogeneous using Eq (i)
∴135x2 + 432xy + 320y2 = 0
A tangent having slope of _
to the ellipse
+ 
= 1 intersects the major & minor axes in points A & B respectively. If C is the centre of the ellipse then the area of the triangle ABC is- a)12 sq. units
- b)24 sq. units
- c)36 sq. units
- d)48 sq. units
Correct answer is option 'B'. Can you explain this answer?
A tangent having slope of _ to the ellipse
to the ellipse + = 1 intersects the major & minor axes in points A & B respectively. If C is the centre of the ellipse then the area of the triangle ABC isa)
12 sq. units
b)
24 sq. units
c)
36 sq. units
d)
48 sq. units
|
Preeti Khanna answered |
Since the major axis is along the y-axis.
∴ Equation of tangent is x = my + [b2m2 + a]1/2
Slope of tangent = 1/m = −4/3
⇒ m = −3/4
Hence, equation of tangent is 4x+3y=24 or
x/6 + y/8 = 1
Its intercepts on the axes are 6 and 8.
Area (ΔAOB) = 1/2×6×8
= 24 sq. unit
∴ Equation of tangent is x = my + [b2m2 + a]1/2
Slope of tangent = 1/m = −4/3
⇒ m = −3/4
Hence, equation of tangent is 4x+3y=24 or
x/6 + y/8 = 1
Its intercepts on the axes are 6 and 8.
Area (ΔAOB) = 1/2×6×8
= 24 sq. unit
The line passing through the extremely A of the major axis and extremity B of the minor axis of the ellipse x2 + 9y2 = 9 meets its auxiliary circle at the point M . Then the area of the triangle with vertices at A,M and the origin O is- a)31/10
- b)29/10
- c)21/10
- d)27/10
Correct answer is option 'D'. Can you explain this answer?
The line passing through the extremely A of the major axis and extremity B of the minor axis of the ellipse x2 + 9y2 = 9 meets its auxiliary circle at the point M . Then the area of the triangle with vertices at A,M and the origin O is
a)
31/10
b)
29/10
c)
21/10
d)
27/10
|
|
Shivam Kumar answered |
27/10
A parabola has the origin as its focus and the line x = 2 as the directrix. Then the vertex of the parabola is at [2008]- a)(0, 2)
- b)(1, 0)
- c)(0, 1)
- d)(2, 0)
Correct answer is option 'B'. Can you explain this answer?
A parabola has the origin as its focus and the line x = 2 as the directrix. Then the vertex of the parabola is at [2008]
a)
(0, 2)
b)
(1, 0)
c)
(0, 1)
d)
(2, 0)
|
|
Geetika Shah answered |
Vertex of a parabola is the mid point of focus and the point
where directrix meets the axis of the parabola.
Here focus is O(0, 0) and directxix meets the axis at B(2, 0)
∴ Vertex of the parabola is (1, 0)
Here focus is O(0, 0) and directxix meets the axis at B(2, 0)
∴ Vertex of the parabola is (1, 0)
The length of the semi-latus-rectum of an ellipse is one third of its major axis, its eccentricity would be
- a)
- b)
- c)2/3
- d)
Correct answer is option 'A'. Can you explain this answer?
The length of the semi-latus-rectum of an ellipse is one third of its major axis, its eccentricity would be
a)
b)
c)
2/3
d)
|
|
Suresh Iyer answered |
Correct Answer :- a
Explanation : Semi latus rectum of ellipse = one half the last rectum
b2/a = 1/3*2a
b2 = 2a2/3
b = (2a/3)1/2...........(1)
So, b2/a = a(1-e2)
b2 = a2(1-e2)
Substituting from (1)
2a2/3 = a2(1-e2)
e2 = 1-2/3
e2 = 1/(3)1/2
If maximum distance of any point on the ellipse x2 + 2y2 + 2xy = 1 from its centre be r, then r is equal to- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
If maximum distance of any point on the ellipse x2 + 2y2 + 2xy = 1 from its centre be r, then r is equal to
a)
b)
c)
d)
|
|
Naina Sharma answered |
Here centre of the ellipse is (0,0) Let P (r cos θ, r sinθ) be any point on the given ellipse then r2 cos2θ + 2r2 sin2 θ + 2r2 sinθ cosθ = 1
An ellipse is drawn by taking a diameter of the circle (x – 1)2 + y2 = 1 as its semi-minor axis and a diameter of the circle x2 + (y – 2)2 = 4 is semi-major axis. If the centre of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is : [2012]- a)4x2 + y2 = 4
- b)x2 + 4y2 = 8
- c)4x2 + y2 = 8
- d)x2 + 4y2 = 16
Correct answer is option 'D'. Can you explain this answer?
An ellipse is drawn by taking a diameter of the circle (x – 1)2 + y2 = 1 as its semi-minor axis and a diameter of the circle x2 + (y – 2)2 = 4 is semi-major axis. If the centre of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is : [2012]
a)
4x2 + y2 = 4
b)
x2 + 4y2 = 8
c)
4x2 + y2 = 8
d)
x2 + 4y2 = 16
|
Ishanya Singh answered |
If the line x – 1 = 0 is the directrix of the parabola y2 – kx + 8 = 0, then one of the values of k is (2000S)- a)1/8
- b)8
- c)4
- d)1/4
Correct answer is option 'C'. Can you explain this answer?
If the line x – 1 = 0 is the directrix of the parabola y2 – kx + 8 = 0, then one of the values of k is (2000S)
a)
1/8
b)
8
c)
4
d)
1/4
|
|
Geetika Shah answered |
KEY CONCEPT : The directrix of the parabola y2 = 4a (x – x1) is given by x = x1 – a.
y2 = kx – 8 ⇒ y2 =
Directrix of parabola is x =
Now, x = 1 also coincides with =
On comparision, =1, or k2 - 4k - 32 = 0
On solving we get k = 4
=1, or k2 - 4k - 32 = 0On solving we get k = 4
The eccentricity of an ellipse, with its centre at the origin, is 1/2. If one of the directrix is x = 4, then the equation of the ellipse is:- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
The eccentricity of an ellipse, with its centre at the origin, is 1/2. If one of the directrix is x = 4, then the equation of the ellipse is:
a)
b)
c)
d)
|
|
Naina Sharma answered |
Let the equation of ellipse (x2)/(a2)+(y2)/(b2)=1
Here a > ba > b because the directrix is parallel to y axis.
b2=a2(1−e2)
Given e= 1/2
⇒b2 = (3/4)a2
But a/e=4
⇒a=2
Putting a=2 we get b= (3)1/2
Required ellipse is (x2)/4+(y2)/3=1
⇒3x2+4y2=12
Here a > ba > b because the directrix is parallel to y axis.
b2=a2(1−e2)
Given e= 1/2
⇒b2 = (3/4)a2
But a/e=4
⇒a=2
Putting a=2 we get b= (3)1/2
Required ellipse is (x2)/4+(y2)/3=1
⇒3x2+4y2=12
The area of the rectangle formed by the perpendiculars from the centre of the standard ellipse to the tangent and normal at its point whose eccentric angle is p/4 is- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
The area of the rectangle formed by the perpendiculars from the centre of the standard ellipse to the tangent and normal at its point whose eccentric angle is p/4 is
a)
b)
c)
d)
|
Lohit Matani answered |
Let equation of ellipse is
x2/a2 + y2/b2 = 1
Equation of tangent at P
(acos π/4, bsin π/4) is x/a + y/b = √2
Equation of normal at P is
√2ax − √2by = a2 − b2
Now
OT = |-√2ab/√(a2 + b2)|
And ON = −(a2 − b2)/[√2 * √(a2 + b2)]
Hence area is (a2 − b2)ab/(a2 + b2)
x2/a2 + y2/b2 = 1
Equation of tangent at P
(acos π/4, bsin π/4) is x/a + y/b = √2
Equation of normal at P is
√2ax − √2by = a2 − b2
Now
OT = |-√2ab/√(a2 + b2)|
And ON = −(a2 − b2)/[√2 * √(a2 + b2)]
Hence area is (a2 − b2)ab/(a2 + b2)
In an ellipse, the distance between its foci is 6 and minor axis is 8. Then its eccentricity is [2006]- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
In an ellipse, the distance between its foci is 6 and minor axis is 8. Then its eccentricity is [2006]
a)
b)
c)
d)
|
Tejas Verma answered |
2ae =6 ⇒ ae = 3 ; 2b =8 ⇒ b = 4
b2 = a2(1-e2) ;16 = a2- a2e2 ⇒ a2 = 16 + 9= 25
⇒ a = 5 
=1, the length of the major axis is
such that the line joining the foci subtends a right angle only at two points on ellipse, is
The locus of the vertices of the family of parabol as 
[2006]- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
The locus of the vertices of the family of parabol as [2006]
[2006]a)
b)
c)
d)
|
Rohit Jain answered |
Given parabola is 
∴ Vertex of parabola is 
To find locus of this vertex,
64xy = 105
which is the required locus.The foci of the ellipse 25 (x + 1)2 + 9(y + 2)2 = 225 are at:- a)(–2, 1) and (–2, 6)
- b)(–1, 2) and (–1, –6)
- c)(–1, –2) and (–1, –6)
- d)(–1, –2) and (–2, –1)
Correct answer is option 'B'. Can you explain this answer?
The foci of the ellipse 25 (x + 1)2 + 9(y + 2)2 = 225 are at:
a)
(–2, 1) and (–2, 6)
b)
(–1, 2) and (–1, –6)
c)
(–1, –2) and (–1, –6)
d)
(–1, –2) and (–2, –1)
|
Athul Roy answered |
Given:
The equation of the ellipse is 25(x-1)^2 + 9(y-2)^2 = 225.
To find:
The foci of the ellipse.
Solution:
Step 1: Identify the standard form of the ellipse equation.
The given equation is in the standard form of an ellipse:
((x-h)^2)/a^2 + ((y-k)^2)/b^2 = 1,
where (h,k) represents the center of the ellipse.
Step 2: Rewrite the given equation in the standard form.
25(x-1)^2 + 9(y-2)^2 = 225
Dividing both sides by 225, we get:
((x-1)^2)/9 + ((y-2)^2)/25 = 1
Comparing this equation with the standard form, we can see that:
a^2 = 9, which means a = 3,
b^2 = 25, which means b = 5.
Step 3: Find the coordinates of the center.
From the given equation, we can see that the center of the ellipse is at (1, 2).
Step 4: Find the coordinates of the foci.
The distance from the center to each focus is given by c, where c^2 = a^2 - b^2.
Plugging in the values of a and b, we have:
c^2 = 9 - 25 = -16
Since c^2 is negative, it means that the ellipse is a vertically elongated ellipse. The foci lie on the major axis, which is the y-axis.
The coordinates of the foci are (h, k + c) and (h, k - c).
Plugging in the values of h, k, and c, we have:
F1: (1, 2 + 4) = (1, 6)
F2: (1, 2 - 4) = (1, -2)
Step 5: Final Answer
The foci of the ellipse are at (1, 6) and (1, -2).
Therefore, the correct answer is option 'B': (1, 2) and (1, 6).
The equation of the ellipse is 25(x-1)^2 + 9(y-2)^2 = 225.
To find:
The foci of the ellipse.
Solution:
Step 1: Identify the standard form of the ellipse equation.
The given equation is in the standard form of an ellipse:
((x-h)^2)/a^2 + ((y-k)^2)/b^2 = 1,
where (h,k) represents the center of the ellipse.
Step 2: Rewrite the given equation in the standard form.
25(x-1)^2 + 9(y-2)^2 = 225
Dividing both sides by 225, we get:
((x-1)^2)/9 + ((y-2)^2)/25 = 1
Comparing this equation with the standard form, we can see that:
a^2 = 9, which means a = 3,
b^2 = 25, which means b = 5.
Step 3: Find the coordinates of the center.
From the given equation, we can see that the center of the ellipse is at (1, 2).
Step 4: Find the coordinates of the foci.
The distance from the center to each focus is given by c, where c^2 = a^2 - b^2.
Plugging in the values of a and b, we have:
c^2 = 9 - 25 = -16
Since c^2 is negative, it means that the ellipse is a vertically elongated ellipse. The foci lie on the major axis, which is the y-axis.
The coordinates of the foci are (h, k + c) and (h, k - c).
Plugging in the values of h, k, and c, we have:
F1: (1, 2 + 4) = (1, 6)
F2: (1, 2 - 4) = (1, -2)
Step 5: Final Answer
The foci of the ellipse are at (1, 6) and (1, -2).
Therefore, the correct answer is option 'B': (1, 2) and (1, 6).
The eccentricity of the ellipse 9x2 + 5y2 – 30y = 0 is:- a)1/3
- b)2/3
- c)3/4
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
The eccentricity of the ellipse 9x2 + 5y2 – 30y = 0 is:
a)
1/3
b)
2/3
c)
3/4
d)
None of these
|
|
Preeti Iyer answered |
9x2 + 5y2 - 30y = 0
9x2 + 5(y−3)2 = 45
We can write it as : [(x-0)2]/5 + [(y-3)2]/9 = 1
Compare it with x2/a2 + y2/b2 = 1
e = [(b2 - a2)/b2]½
e = [(9-5)/9]1/2
e = (4/9)½
e = ⅔
9x2 + 5(y−3)2 = 45
We can write it as : [(x-0)2]/5 + [(y-3)2]/9 = 1
Compare it with x2/a2 + y2/b2 = 1
e = [(b2 - a2)/b2]½
e = [(9-5)/9]1/2
e = (4/9)½
e = ⅔
x – 2y + 4 = 0 is a common tangent to y2 = 4x & 
+ 
= 1. Then the value of b and the other common tangent are given by- a)b = √3 ;x + 2y + 4 = 0
- b)b = 3; x + 2y + 4 = 0
- c)b = √3 ;x + 2y– 4 = 0
- d)b = √3 ; x – 2y – 4 = 0
Correct answer is option 'A'. Can you explain this answer?
x – 2y + 4 = 0 is a common tangent to y2 = 4x & + = 1. Then the value of b and the other common tangent are given by
+ = 1. Then the value of b and the other common tangent are given bya)
b = √3 ;x + 2y + 4 = 0
b)
b = 3; x + 2y + 4 = 0
c)
b = √3 ;x + 2y– 4 = 0
d)
b = √3 ; x – 2y – 4 = 0
|
Knowledge Hub answered |
Equation of tangent of ellipse is
y=mx±+ ..............(i)
Given equation is x−2y+4=0 .........(ii)
Since (i) & (ii) are same, comparing them, we get
m=1/2&a2m2 + b2 = 2
⇒4*1/4+b2=4
b=±3
Equation of tangent of parabola
y=mx+1/m .........(iii)
by (i) & (iii)
1/m^2= a2m2+b2
on solving it we get m=±1/2
with m = −1/2 , we get x+2y+4=0
y=mx±+ ..............(i)
Given equation is x−2y+4=0 .........(ii)
Since (i) & (ii) are same, comparing them, we get
m=1/2&a2m2 + b2 = 2
⇒4*1/4+b2=4
b=±3
Equation of tangent of parabola
y=mx+1/m .........(iii)
by (i) & (iii)
1/m^2= a2m2+b2
on solving it we get m=±1/2
with m = −1/2 , we get x+2y+4=0
In the ellipse x2 + 3y2 = 9 the distance between the foci is- a)3
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
In the ellipse x2 + 3y2 = 9 the distance between the foci is
a)
3
b)
c)
d)
|
Learners Habitat answered |
x2 + 3y2 = 9
⇒(x2)/9 + (y2)/3=1
⇒a2=9, b2=3
⇒e=[1−b2/a2]1/2
=(2/3)1/2
Therefore, distance between foci is =2ae = 2 × 3 × (2/3)1/2
=2(6)1/2
⇒(x2)/9 + (y2)/3=1
⇒a2=9, b2=3
⇒e=[1−b2/a2]1/2
=(2/3)1/2
Therefore, distance between foci is =2ae = 2 × 3 × (2/3)1/2
=2(6)1/2
The number of points on X-axis which are at a distance c units (c < 3) from (2, 3) is- a)1
- b)0
- c)3
- d)2
Correct answer is option 'B'. Can you explain this answer?
The number of points on X-axis which are at a distance c units (c < 3) from (2, 3) is
a)
1
b)
0
c)
3
d)
2
|
Mohit Rajpoot answered |
Distance of 'c' units from (2,3)
Let the no: of points be (x,0)
By distance formula
{(2−x)2+(3−0)2}=c
4−4x+x2+9=c
⇒x2−4x+13 = c:c=2,2
There are the points of c,such that when they are applied back to the equations,the number of points will become zero.
Let the no: of points be (x,0)
By distance formula
{(2−x)2+(3−0)2}=c
4−4x+x2+9=c
⇒x2−4x+13 = c:c=2,2
There are the points of c,such that when they are applied back to the equations,the number of points will become zero.
PQ is a double ordinate of the ellipse x2+ 9y2 = 9, the normal at P meets the diameter through Q at R, then the locus of the mid point of PR is- a)A circle
- b)A parabola
- c)An ellipse
- d)A hyperbola
Correct answer is option 'C'. Can you explain this answer?
PQ is a double ordinate of the ellipse x2+ 9y2 = 9, the normal at P meets the diameter through Q at R, then the locus of the mid point of PR is
a)
A circle
b)
A parabola
c)
An ellipse
d)
A hyperbola
|
Avik Nair answered |
Given an ellipse with equation x^2/9 + y^2/1 = 1, let's analyze the given information step by step to understand why the locus of the midpoint of PR is an ellipse.
1. Finding the coordinates of point P:
Since PQ is a double ordinate, it means that the line segment PQ passes through the center of the ellipse. The center of the ellipse is (0, 0), so the coordinates of point P are (0, 1).
2. Finding the equation of the normal at point P:
To find the equation of the normal at point P, we need to find the slope of the tangent at point P. Differentiating the equation of the ellipse, we get:
2x/9 + 2y(dy/dx) = 0
dy/dx = -9x/(2y)
The slope of the normal is the negative reciprocal of the slope of the tangent, so the slope of the normal at P is (2y)/(9x).
Using the point-slope form of a line, the equation of the normal at P is given by:
y - 1 = (2y)/(9x) * (x - 0)
9xy - 9x = 2y^2 - 2y
2y^2 - 9xy + 9x - 2y + 2 = 0
3. Finding the coordinates of point Q:
Since PQ is a double ordinate, it means that the distance between points P and Q is equal to 2 times the distance between point P and the center of the ellipse. The distance between P and the center is 1, so the distance between P and Q is 2.
Since point P is at (0, 1), point Q can be at either (2, 1) or (-2, 1).
4. Finding the equation of the diameter through point Q:
Since point Q lies on the diameter, we can find the equation of the diameter passing through point Q by finding the equation of the line passing through points P and Q.
Using the two-point form of a line, the equation of the diameter is given by:
(y - 1) = (1 - 1)/(0 - 2) * (x - 2)
(x + 2) = 0
So the equation of the diameter is x = -2.
5. Finding the coordinates of point R:
Point R is the intersection of the normal at P and the diameter passing through Q. Since the equation of the diameter is x = -2, the x-coordinate of point R is -2. We can substitute this value into the equation of the normal to find the y-coordinate of point R.
2y^2 - 9(-2)y + 9(-2) - 2y + 2 = 0
2y^2 + 18y - 18 - 2y + 2 = 0
2y^2 + 16y - 16 = 0
y^2 + 8y - 8 = 0
Using the quadratic formula, we can solve for y:
y = (-8 ± √(8^2 - 4*1*(-8)))/2
y = (-8 ± √(64 + 32))/2
y = (-8 ± √96)/2
y = (-
1. Finding the coordinates of point P:
Since PQ is a double ordinate, it means that the line segment PQ passes through the center of the ellipse. The center of the ellipse is (0, 0), so the coordinates of point P are (0, 1).
2. Finding the equation of the normal at point P:
To find the equation of the normal at point P, we need to find the slope of the tangent at point P. Differentiating the equation of the ellipse, we get:
2x/9 + 2y(dy/dx) = 0
dy/dx = -9x/(2y)
The slope of the normal is the negative reciprocal of the slope of the tangent, so the slope of the normal at P is (2y)/(9x).
Using the point-slope form of a line, the equation of the normal at P is given by:
y - 1 = (2y)/(9x) * (x - 0)
9xy - 9x = 2y^2 - 2y
2y^2 - 9xy + 9x - 2y + 2 = 0
3. Finding the coordinates of point Q:
Since PQ is a double ordinate, it means that the distance between points P and Q is equal to 2 times the distance between point P and the center of the ellipse. The distance between P and the center is 1, so the distance between P and Q is 2.
Since point P is at (0, 1), point Q can be at either (2, 1) or (-2, 1).
4. Finding the equation of the diameter through point Q:
Since point Q lies on the diameter, we can find the equation of the diameter passing through point Q by finding the equation of the line passing through points P and Q.
Using the two-point form of a line, the equation of the diameter is given by:
(y - 1) = (1 - 1)/(0 - 2) * (x - 2)
(x + 2) = 0
So the equation of the diameter is x = -2.
5. Finding the coordinates of point R:
Point R is the intersection of the normal at P and the diameter passing through Q. Since the equation of the diameter is x = -2, the x-coordinate of point R is -2. We can substitute this value into the equation of the normal to find the y-coordinate of point R.
2y^2 - 9(-2)y + 9(-2) - 2y + 2 = 0
2y^2 + 18y - 18 - 2y + 2 = 0
2y^2 + 16y - 16 = 0
y^2 + 8y - 8 = 0
Using the quadratic formula, we can solve for y:
y = (-8 ± √(8^2 - 4*1*(-8)))/2
y = (-8 ± √(64 + 32))/2
y = (-8 ± √96)/2
y = (-
Q is a point on the auxiliary circle of an ellipse. P is the corresponding point on ellipse. N is the foot of perpendicular from focus S, to the tangent of auxiliary circle at Q. Then- a)SP = SN
- b)SP = PQ
- c)PN = SP
- d)NQ = SP
Correct answer is option 'A'. Can you explain this answer?
Q is a point on the auxiliary circle of an ellipse. P is the corresponding point on ellipse. N is the foot of perpendicular from focus S, to the tangent of auxiliary circle at Q. Then
a)
SP = SN
b)
SP = PQ
c)
PN = SP
d)
NQ = SP
|
|
Naina Sharma answered |
P = (acosθ, bsinθ)
Q = (acosθ, asinθ)
equation of tangent at Q
Q = (acosθ, asinθ)
equation of tangent at Q
The ellipse x2 + 4y2 = 4 is inscribed in a rectangle aligned with the coordinate axes, which in turn is inscribed in another ellipse that passes through the point (4, 0). Then the equation of the ellipse is : [2009]- a)x2 + 12y2= 16
- b)4 x2 + 48y2= 48
- c)4 x2 + 64y2 = 48
- d)x2 + 16y2= 16
Correct answer is option 'A'. Can you explain this answer?
The ellipse x2 + 4y2 = 4 is inscribed in a rectangle aligned with the coordinate axes, which in turn is inscribed in another ellipse that passes through the point (4, 0). Then the equation of the ellipse is : [2009]
a)
x2 + 12y2= 16
b)
4 x2 + 48y2= 48
c)
4 x2 + 64y2 = 48
d)
x2 + 16y2= 16
|
Anushka Ahuja answered |
Solution:
Given that the ellipse x^2/4 + y^2 = 1 is inscribed in a rectangle aligned with the coordinate axes and this rectangle is inscribed in another ellipse that passes through the point (4, 0).
Let's find the equation of the outer ellipse.
Step 1: Find the coordinates of the four corners of the rectangle.
The corners of the rectangle can be obtained by solving the equation of the ellipse with x = ±2 and y = ±1.
For x = 2, we have 4/4 + y^2 = 1, which gives y = ±√3.
So, the coordinates are (2, √3) and (2, -√3).
Similarly, for x = -2, we have 4/4 + y^2 = 1, which gives y = ±√3.
So, the coordinates are (-2, √3) and (-2, -√3).
Step 2: Find the equation of the ellipse passing through the points (2, √3), (2, -√3), (-2, √3), and (-2, -√3).
Using the standard form of an ellipse, the equation is:
(x - h)^2/a^2 + (y - k)^2/b^2 = 1, where (h, k) is the center of the ellipse, and a and b are the semi-major and semi-minor axes, respectively.
Step 3: Find the center of the ellipse.
The center of the ellipse is the midpoint of the diagonals of the rectangle.
The midpoint of the diagonals is ((2 + (-2))/2, (√3 + (-√3))/2) = (0, 0).
Step 4: Find the semi-major and semi-minor axes.
The semi-major axis is the distance from the center to one of the corners of the rectangle, which is 2.
The semi-minor axis is the distance from the center to one of the sides of the rectangle, which is 1.
Therefore, the equation of the outer ellipse is:
(x - 0)^2/2^2 + (y - 0)^2/1^2 = 1
Simplifying, we get x^2/4 + y^2 = 1.
Hence, the correct answer is option A) x^2 - 12y^2 = 16.
Given that the ellipse x^2/4 + y^2 = 1 is inscribed in a rectangle aligned with the coordinate axes and this rectangle is inscribed in another ellipse that passes through the point (4, 0).
Let's find the equation of the outer ellipse.
Step 1: Find the coordinates of the four corners of the rectangle.
The corners of the rectangle can be obtained by solving the equation of the ellipse with x = ±2 and y = ±1.
For x = 2, we have 4/4 + y^2 = 1, which gives y = ±√3.
So, the coordinates are (2, √3) and (2, -√3).
Similarly, for x = -2, we have 4/4 + y^2 = 1, which gives y = ±√3.
So, the coordinates are (-2, √3) and (-2, -√3).
Step 2: Find the equation of the ellipse passing through the points (2, √3), (2, -√3), (-2, √3), and (-2, -√3).
Using the standard form of an ellipse, the equation is:
(x - h)^2/a^2 + (y - k)^2/b^2 = 1, where (h, k) is the center of the ellipse, and a and b are the semi-major and semi-minor axes, respectively.
Step 3: Find the center of the ellipse.
The center of the ellipse is the midpoint of the diagonals of the rectangle.
The midpoint of the diagonals is ((2 + (-2))/2, (√3 + (-√3))/2) = (0, 0).
Step 4: Find the semi-major and semi-minor axes.
The semi-major axis is the distance from the center to one of the corners of the rectangle, which is 2.
The semi-minor axis is the distance from the center to one of the sides of the rectangle, which is 1.
Therefore, the equation of the outer ellipse is:
(x - 0)^2/2^2 + (y - 0)^2/1^2 = 1
Simplifying, we get x^2/4 + y^2 = 1.
Hence, the correct answer is option A) x^2 - 12y^2 = 16.
If a tangent of slope 2 of the ellipse 
is normal to the circle x2 + y2 + 4x + 1 = 0 , then the maximum value of ab is- a)4
- b)2
- c)1
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
If a tangent of slope 2 of the ellipse is normal to the circle x2 + y2 + 4x + 1 = 0 , then the maximum value of ab is
is normal to the circle x2 + y2 + 4x + 1 = 0 , then the maximum value of ab isa)
4
b)
2
c)
1
d)
None of these
|
|
Geetika Shah answered |
Let 
be the tangent
It is passing through(-2,0)
be the tangentIt is passing through(-2,0)
Let (x, y) be any point on the parabola y2 = 4x. Let P be the point that divides the line segment from (0, 0) to (x, y) in the ratio 1 : 3. Then the locus of P is (2011)- a)x2 = y
- b)y2 = 2x
- c)y2 = x
- d)x2 = 2y
Correct answer is option 'C'. Can you explain this answer?
Let (x, y) be any point on the parabola y2 = 4x. Let P be the point that divides the line segment from (0, 0) to (x, y) in the ratio 1 : 3. Then the locus of P is (2011)
a)
x2 = y
b)
y2 = 2x
c)
y2 = x
d)
x2 = 2y
|
|
Geetika Shah answered |
Let A (x, y) = (t2, 2 t) be any point on parabola y2 = 4x.
Let P (h, k) divides OA in the ratio 1 : 3
Then (h, k) = 
∴ locus of P (h, k) is x = y2.
A circle has the same centre as an ellipse & passes through the foci F1 & F2 of the ellipse, such that the two curves intersect in 4 points. Let `P' be any one of their point of intersection. If the major axis of the ellipse is 17 & the area of the triangle PF1F2 is 30, then the distance between the foci is less than- a)12
- b)13
- c)14
- d)15
Correct answer is option 'B'. Can you explain this answer?
A circle has the same centre as an ellipse & passes through the foci F1 & F2 of the ellipse, such that the two curves intersect in 4 points. Let `P' be any one of their point of intersection. If the major axis of the ellipse is 17 & the area of the triangle PF1F2 is 30, then the distance between the foci is less than
a)
12
b)
13
c)
14
d)
15
|
|
Tanuja Kapoor answered |
The normal at a variable point P on an ellipse 
+ 
= 1 of eccentricity e meets the axes of the ellipse in Q and R then the locus of the mid-point of QR is a conic with an eccentricity e' such that- a)e' is independent of e
- b)e' = 1
- c)e' = e
- d)e' = 1/e
Correct answer is option 'C'. Can you explain this answer?
The normal at a variable point P on an ellipse + = 1 of eccentricity e meets the axes of the ellipse in Q and R then the locus of the mid-point of QR is a conic with an eccentricity e' such that
+ = 1 of eccentricity e meets the axes of the ellipse in Q and R then the locus of the mid-point of QR is a conic with an eccentricity e' such thata)
e' is independent of e
b)
e' = 1
c)
e' = e
d)
e' = 1/e
|
|
Pooja Shah answered |
e’ = e
An ellipse is such that the length of the latus rectum is equal to the sum of the lengths of its semi principal axes. Then- a)Ellipse becomes a circle
- b)Ellipse becomes a line segment between the two foci
- c)Ellipse becomes a parabola
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
An ellipse is such that the length of the latus rectum is equal to the sum of the lengths of its semi principal axes. Then
a)
Ellipse becomes a circle
b)
Ellipse becomes a line segment between the two foci
c)
Ellipse becomes a parabola
d)
None of these
|
|
Pooja Shah answered |
Sum of the lengths of its semi−principal axis is : a+b
Given, 2b2/a=a+b
Or,2b2=a2+ab
Or,2b2−ab−a2 =0
Or,2b2−2ab+ab−a2=0
Or,2b(b−a)+a(b−a)=0
Or,(2b+a)(b−a)=0
Since,both are positive quantity, b=a
∴ It’s a circle.
Given, 2b2/a=a+b
Or,2b2=a2+ab
Or,2b2−ab−a2 =0
Or,2b2−2ab+ab−a2=0
Or,2b(b−a)+a(b−a)=0
Or,(2b+a)(b−a)=0
Since,both are positive quantity, b=a
∴ It’s a circle.
If tan q1. tan q2 = –
then the chord joining two points q1 & q2 on the ellipse 
+ 
= 1 will subtend a right angle at- a)Focus
- b)Centre
- c)End of the major axis
- d)End of the minor axis
Correct answer is option 'B'. Can you explain this answer?
If tan q1. tan q2 = – then the chord joining two points q1 & q2 on the ellipse + = 1 will subtend a right angle at
then the chord joining two points q1 & q2 on the ellipse + = 1 will subtend a right angle ata)
Focus
b)
Centre
c)
End of the major axis
d)
End of the minor axis
|
Lavanya Menon answered |
Clearly, they subtend right angle at centre.
If x + y = k is normal to y2 = 12 x, then k is (2000S)- a)3
- b)9
- c)–9
- d)–3
Correct answer is option 'B'. Can you explain this answer?
If x + y = k is normal to y2 = 12 x, then k is (2000S)
a)
3
b)
9
c)
–9
d)
–3
|
Gaurav Rane answered |
y = mx + c is normal to the parabola y2 = 4 ax if c = – 2am – am3
Here m = –1, c = k and a = 3
∴ c = k = – 2 (3) (–1) – 3 (–1)3 = 9
Here m = –1, c = k and a = 3
∴ c = k = – 2 (3) (–1) – 3 (–1)3 = 9
An ellipse has the points (1, -1) and (2, -1) as its foci and x +y - 5 = 0 as one of its tangents. Then the point where this line touches the ellipse is- a)
- b)
- c)
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
An ellipse has the points (1, -1) and (2, -1) as its foci and x +y - 5 = 0 as one of its tangents. Then the point where this line touches the ellipse is
a)
b)
c)
d)
None of these
|
Nabanita Singh answered |
Let image of S'' be with respect to x +y - 5 = 0
Let P be the point of contact.
Because the line L = 0 is tangent to the ellipse, there exists a point P uniquely on the line such that PS + PS ' = 2a .
Since PS ' = 2a Hence, P should be the collinear with SS ''
Hence P is a point of intersection of SS '' (4x - 5 y = 9) , and
Let P be the point of contact.
Because the line L = 0 is tangent to the ellipse, there exists a point P uniquely on the line such that PS + PS ' = 2a .
Since PS ' = 2a Hence, P should be the collinear with SS ''
Hence P is a point of intersection of SS '' (4x - 5 y = 9) , and
Two common tangents to the circle x2 + y2 = 2a2 and parabola y2 = 8ax are [2002]- a)x = ±( y+ 2a)
- b)y = ± ( x+ 2a)
- c)x = ± ( y+a)
- d)y = ± ( x+a)
Correct answer is option 'B'. Can you explain this answer?
Two common tangents to the circle x2 + y2 = 2a2 and parabola y2 = 8ax are [2002]
a)
x = ±( y+ 2a)
b)
y = ± ( x+ 2a)
c)
x = ± ( y+a)
d)
y = ± ( x+a)
|
|
Juhi Nambiar answered |
Any tangent to the parabola y2 = 8ax is
...(i)
...(i)If (i) is a tangent to the circle, x2 + y2 = 2a2 then,
⇒ m2(1 + m2) = 2 ⇒ (m2 + 2)(m2 – 1) = 0 ⇒ m = ± 1.
So from (i), y = ± (x + 2a).
So from (i), y = ± (x + 2a).
+ 
= 1 at the positive end of latus rectum is
An ellipse having foci at (3,3) and (-4,4) and passing though the origin has eccentricity equal to - a)3/7
- b)2/7
- c)5/7
- d)3/5
Correct answer is option 'C'. Can you explain this answer?
An ellipse having foci at (3,3) and (-4,4) and passing though the origin has eccentricity equal to
a)
3/7
b)
2/7
c)
5/7
d)
3/5
|
Sanchita Reddy answered |
Ellipse passing through O (0, 0) and having foci P (3, 3) and Q (-4, 4) ,
Then
Then
The locus of the orthocentre of the triangle formed by the lines
(1+ p) x – py + p (1+ p) = 0,
(1+ q) x – qy + q (1+ q) = 0,
and y = 0, where p ≠ q, is (2009)- a)a hyperbola
- b)a parabola
- c)an ellipse
- d)a straight line
Correct answer is option 'D'. Can you explain this answer?
The locus of the orthocentre of the triangle formed by the lines
(1+ p) x – py + p (1+ p) = 0,
(1+ q) x – qy + q (1+ q) = 0,
and y = 0, where p ≠ q, is (2009)
(1+ p) x – py + p (1+ p) = 0,
(1+ q) x – qy + q (1+ q) = 0,
and y = 0, where p ≠ q, is (2009)
a)
a hyperbola
b)
a parabola
c)
an ellipse
d)
a straight line
|
Saranya Choudhary answered |
The triangle is formed by the lines
AB : (1 + p) x - py + p (1 +p)= 0
AC : (1+ q)x -qy +q(1+ q)= 0
BC :y =0
So that the vertices are A( pq, ( p + 1)(q+ 1)), B(- p, 0), C (-q, 0)
Let H (h,k) be the orthocentre of Δ ABC.
Then as AH ^ BC and passes through A( pq, ( p + 1)(q+ 1)) The eqn of AH is x = pq
∴ h = pq ...(1)
Also BH is perpendicular to AC
AC : (1+ q)x -qy +q(1+ q)= 0
BC :y =0
So that the vertices are A( pq, ( p + 1)(q+ 1)), B(- p, 0), C (-q, 0)
Let H (h,k) be the orthocentre of Δ ABC.
Then as AH ^ BC and passes through A( pq, ( p + 1)(q+ 1)) The eqn of AH is x = pq
∴ h = pq ...(1)
Also BH is perpendicular to AC
(using eqn (1)⇒ k =- pq ...(2)
From (1) and (2) we observe h + k = 0
∴ Locus of (h, k) is x + y = 0 which is a straight line.
The tangent at the point α on the ellipse 
meets the auxiliary circle in two points which subtends a right angle at the centre, then the eccentricity ‘e’ of the ellipse is given by the equation :- a)e2 (1 + cos2 α) = 1
- b)e2 (cosec2α - 1) = 1
- c)e2 (1 + sin2 α) = 1
- d)e2 (1 + tan2 α) = 1
Correct answer is option 'C'. Can you explain this answer?
The tangent at the point α on the ellipse meets the auxiliary circle in two points which subtends a right angle at the centre, then the eccentricity ‘e’ of the ellipse is given by the equation :
meets the auxiliary circle in two points which subtends a right angle at the centre, then the eccentricity ‘e’ of the ellipse is given by the equation :a)
e2 (1 + cos2 α) = 1
b)
e2 (cosec2α - 1) = 1
c)
e2 (1 + sin2 α) = 1
d)
e2 (1 + tan2 α) = 1
|
Sinjini Datta answered |
Method to Solve :
Let P be the point on the parabola, y2 = 8x which is at a minimum distance from the centre C of the circle, x2 + (y + 6)2 = 1.Then the equation of the circle, passing through C and having its centre at P is: [JEE M 2016]- a)
- b)x2 + y2 – 4x + 9y + 18 = 0
- c)x2 + y2 – 4x + 8y + 12 = 0
- d)x2 + y2 – x + 4y – 12 = 0
Correct answer is option 'C'. Can you explain this answer?
Let P be the point on the parabola, y2 = 8x which is at a minimum distance from the centre C of the circle, x2 + (y + 6)2 = 1.Then the equation of the circle, passing through C and having its centre at P is: [JEE M 2016]
a)
b)
x2 + y2 – 4x + 9y + 18 = 0
c)
x2 + y2 – 4x + 8y + 12 = 0
d)
x2 + y2 – x + 4y – 12 = 0
|
|
Vicky Kumar answered |
t ∈ R represents
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