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Test: Ellipse- 4 - JEE MCQ


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30 Questions MCQ Test Chapter-wise Tests for JEE Main & Advanced - Test: Ellipse- 4

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Test: Ellipse- 4 - Question 1

An ellipse having foci at (3,3) and (–4,4) and passing though the origin has eccentricity equal to 

Detailed Solution for Test: Ellipse- 4 - Question 1

Ellipse passing through O (0, 0) and having foci P (3, 3) and Q (-4, 4),

Then 

Test: Ellipse- 4 - Question 2

The length of the major axis of the ellipse (5x - 10)2 + (5y + 15)2 

Detailed Solution for Test: Ellipse- 4 - Question 2

(5x - 10)2 + (5y + 15)2 
⇒ (x - 2)2 + (y +3)2 


 is an ellipse, whose focus is (2, -3) , directrix
3x - 4y + 7 = 0 and eccentricity is 1/2.
Length of from focus to directric is 



So length of major axis is 20/3

Test: Ellipse- 4 - Question 3

The eccentric angle of a point on the ellipse  at a distance of 5/4 units from the focus on the positive x-axis, is 

Detailed Solution for Test: Ellipse- 4 - Question 3

Any point on the ellipse is  . Thae focus on the positive x-axis is (1,0) .
Given that

Test: Ellipse- 4 - Question 4

From any point P lying in first quadrant on the ellipse  PN is drawn perpendicular to the major axis and produced at Q so that NQ equals to PS, where S is a focus. Then the locus of Q is 

Detailed Solution for Test: Ellipse- 4 - Question 4



Let point Q be (h,k) , where k < 0
Given that k = SP = a + ex1 , where P (x1, y1) lies on the ellipse

⇒ -y = a + ex
⇒ 3x + 5y + 25 = 0

Test: Ellipse- 4 - Question 5

If a tangent of slope 2 of the ellipse  normal to the circle x2 + y2 + 4x + 1 = 0 , then the maximum value of ab is

Detailed Solution for Test: Ellipse- 4 - Question 5

Let y =  be the tangent
It is passing through(-2,0) 

      (AM > GM)
∴ ab < 4

Test: Ellipse- 4 - Question 6

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

Detailed Solution for Test: Ellipse- 4 - Question 6


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 - 5y= 9) , and x + y - 5 = 0 i.e.

Test: Ellipse- 4 - Question 7

From point P (8, 27), tangent PQ and PR are drawn to the ellipse  Then the angle subtended by QR at origin is 

Detailed Solution for Test: Ellipse- 4 - Question 7

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

Test: Ellipse- 4 - Question 8

An ellipse is sliding along the co-ordinate axes. If the foci of the ellipse are (1,1) and (3,3), then area of the director circle of the ellipse (in sq. units) is

Detailed Solution for Test: Ellipse- 4 - Question 8

Since x-axis and y-axis are perpendicular tangents to the ellipse, (0,0) lies on the director circle and midpoint of foci (2,2) is centre of the circle. 
Hence, radius = 2√2
⇒ the area is 8π units.

Test: Ellipse- 4 - Question 9

Number of point on the ellipse   from which pair of perpendicular tangents are drawn to the ellipse 

Detailed Solution for Test: Ellipse- 4 - Question 9

For the ellipse  
Equation of director circle is x2 + y2 = 25. The director circle will cut the ellipse

Test: Ellipse- 4 - Question 10

The equation of the ellipse whose axes are coincident with the co-ordinates axes and which touches the straight lines 3x - 2y - 20 = 0 and x + 6y - 20 = 0 is 

Detailed Solution for Test: Ellipse- 4 - Question 10

Let the equation of the ellipse be 

We know that the general equation of the tangent to the ellipse is
 .......(i)
Since 2x - 2y - 20 = 0 or  is tangent to the ellipse .
Comparing with eq (i)

      ......(ii)
Similarly, since x + 6y - 20 = 0 i.e, is tangent to the ellipse, therefore 
comparing with eq. …(i)

⇒ a2 + 36b2 = 400     ......(iii)
Solving eqs. (ii) and (iii) , we get a2 = 40 and b2 = 10 , Therefore , the required equation of the ellipse is

Test: Ellipse- 4 - Question 11

If maximum distance of any point on the ellipse x2 + 2y2 + 2xy = 1 from its centre be r, then r is equal to

Detailed Solution for Test: Ellipse- 4 - Question 11

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

Test: Ellipse- 4 - Question 12

The equation of the line passing through the centre and bisecting the chord 7x + y -1 = 0 of the ellipse 

Detailed Solution for Test: Ellipse- 4 - Question 12

Let (h,k) be the midpoint of the chord 7x + y -1 = 0
  ......(i)
⇒ 7x + y = 1
Represents same straight line 

⇒ Equation of the line joining (0,0) and (h,k)  is y - x= 0 .

Test: Ellipse- 4 - Question 13

Eccentricity of ellipse  such that the line joining the foci subtends a right angle only at two points on ellipse, is

Test: Ellipse- 4 - Question 14

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 :

Detailed Solution for Test: Ellipse- 4 - Question 14

Equation of auxillary circle is x2 + y2 = a2
Let P is (a cos α, b sinα)
Equation of AB is 
b cos α. x + a sin α y = ab .....(i) 
To get combined equation of CA and CB, homogenize equation of circle with equation (i),

since ∠BCA = 90°
∴ coefficient of x2 + coefficient of y2 =  0

⇒ b2 + b2 - b2 cos2 α - a2 sin2 α = 0
b2(sin2 α+1) = a2 sin2α
⇒ 


Test: Ellipse- 4 - Question 15

The number of values of c such that the straight line y = 4x+ c touches the curve 

Detailed Solution for Test: Ellipse- 4 - Question 15

For given slope there exists two parallel tangents to ellipse. Hence, there are two values of c.

Test: Ellipse- 4 - Question 16

If tangents are drawn to the ellipse x2 + 2y2 = 2, then the locus of the midpoint of the intercept made by the tangent between the coordinate axes is

Detailed Solution for Test: Ellipse- 4 - Question 16

For any tangent to ellipse

is given by 

Using midpoint formula, we have





Test: Ellipse- 4 - Question 17

An ellipse with major and minor axes, 6√3 and 6 respectively slides along the co-ordinate axes and always remains confined in the first quadrant. If the length of arc described by the centre of the ellipse is kπ/6, then the value of k is

Test: Ellipse- 4 - Question 18

Consider the ellipse   and the parabola y2 = 2x. They intersect at P and Q in the first and fourth quadrants respectively. Tangents to the ellipse at P and Q intersect the x-axis at R and tangents to the parabola at P and Q intersect the x-axis at S. If the area of quadrilateral PRQS is λ then 

Detailed Solution for Test: Ellipse- 4 - Question 18

Area of quadrilateral 

Test: Ellipse- 4 - Question 19

The area of the quadrilateral formed by the tangents at the end points of latus rectum of the ellipse  is k then k/9 is

Test: Ellipse- 4 - Question 20

The length of the focal chord of the ellipse   which is inclined to x – axis at an angle 45° is λ, then

Test: Ellipse- 4 - Question 21

Q is a point on the auxiliary circle corresponding to the point P of the ellipse  If T is the foot of the perpendicular dropped from the focus S on to the tangent to the auxiliary circle at Q then the ΔSPT is :

Detailed Solution for Test: Ellipse- 4 - Question 21

Tangent at Q is x cos θ + y sin θ = a

ST = |a e cos θ - a| = a|(1 -e cos θ)
Also SP = e PM 

ST = SP ⇒ isosceles

Test: Ellipse- 4 - Question 22

let x and y satisfy the relation x2 + 9y2 - 4x + 6y + 4 = 0, then maximum value of the expression (4x - 9y),

Detailed Solution for Test: Ellipse- 4 - Question 22

Given equation is

Let x - 2 
Then 4x - 9y = 11 + 4 cos θ - 3sin θ

Test: Ellipse- 4 - Question 23

A rectangle ABCD has area 200 square units. An ellipse with area 200π passes through A and C and has foci at B and D, then

Test: Ellipse- 4 - Question 24

A tangent to the ellipse  at any point P meet the line x = 0 at a point Q. Let R be the image of Q in the line y = x, then circle whose extremities of a diameter are Q and R passes through a fixed point. The fixed point is

Detailed Solution for Test: Ellipse- 4 - Question 24

Equation of the tangent to the ellipse at P (5 cos θ , 4 sin θ) is

It meets the line x = 0 at Q (0, 4 cosec θ)
Image of Q in the line y = x is R (4 cosec θ , 0)
∴ Equation of the circle is
X (x – 4 cosec θ) + y(y – 4 cosec θ) = 0
i.e. x2 + y2 – 4 (x + y) cosec θ = 0
∴ each member of the family passes through the intersection of x2 + y2 = 0 and x + y = 0
i.e. the point (0, 0).

Test: Ellipse- 4 - Question 25

If the curve x2 + 3y2 = 9 subtends an obtuse angle at the point (2α, α), then a possible value of α2 is 

Detailed Solution for Test: Ellipse- 4 - Question 25

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.,


Test: Ellipse- 4 - Question 26

Let S = 0 be the equation of reflection of  about the line x – y – 2 = 0. Then the locus of point of intersection of perpendicular tangents of S is

Detailed Solution for Test: Ellipse- 4 - Question 26

(x - 5)2 + (y - 2)2 = 16 + 9

Test: Ellipse- 4 - Question 27

Any ordinate MP of a ellipse  meets the auxiliary circle in Q, then locus of point of intersection of normals at P and Q to the respective curves, is

Detailed Solution for Test: Ellipse- 4 - Question 27

Equation of normal to the ellipse at ‘P’ is
5x sec θ – 3y cosec θ = 16 …(1)
Equation of normal to the circle x2 + y2 = 25 at point Q is –

y = x tan θ
Eliminating θ from (1) & (2). We get x2 + y2 = 64.

Test: Ellipse- 4 - Question 28

The radius of the largest circle whose centre at (–3, 0) and is inscribed in the ellipse 16x2 + 25y2 = 400 is

Test: Ellipse- 4 - Question 29

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

Detailed Solution for Test: Ellipse- 4 - Question 29


Equation of line AM is
x +3y-3 = 0
Perpendicular distance of line from the origin 

Length of AM 

Test: Ellipse- 4 - Question 30

Coordinate of the vertices B and C of triangle ABC are respectively (2, 0) and  (8, 0). The vertex ‘A’ is varying in such a way that  Then the locus of ‘A’ is an ellipse whose major axis is of length. 

Detailed Solution for Test: Ellipse- 4 - Question 30

Let BC = a, CA = b & AB = c



Locus of A is an ellipse with major axis of length 10

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