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Test: Ellipse- 3 - JEE MCQ


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30 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Test: Ellipse- 3

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Test: Ellipse- 3 - 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- 3 - Question 1

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

Test: Ellipse- 3 - Question 2

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

Detailed Solution for Test: Ellipse- 3 - Question 2






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



So length of major axis is 20/3

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Test: Ellipse- 3 - 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- 3 - Question 3

Any point on the ellipse is (2 cosθ, √3 sinθ). The focus on the positive x-axis is (1,0). 

Given that (2cosθ-1)2 + 3sin2 θ = 25/16
⇒ cos θ = 3/4

Test: Ellipse- 3 - 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- 3 - 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

Test: Ellipse- 3 - Question 5

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

Detailed Solution for Test: Ellipse- 3 - Question 5

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

 

Test: Ellipse- 3 - 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- 3 - 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 - 5 y = 9) , and 

Test: Ellipse- 3 - 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- 3 - 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 + 320 y2 = 0

Test: Ellipse- 3 - 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- 3 - 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- 3 - Question 9

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- 3 - Question 9

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


is tangent to the ellipse .
Comparing with eq (i)


  ......(ii)
Similarly, since
 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- 3 - Question 10

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

Detailed Solution for Test: Ellipse- 3 - Question 10

For the ellipse 
Equation of director circle is x2 +y2 = 25. The director circle will cut the ellipse at 4 points

Test: Ellipse- 3 - 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- 3 - 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 sin2θ + 2r2 sin θ cos θ = 1

Test: Ellipse- 3 - 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- 3 - Question 12

Let (h,k) be the midpoint of the chord 7 x +y -1 = 0

Represents same straight line 

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

Test: Ellipse- 3 - Question 13

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

Detailed Solution for Test: Ellipse- 3 - Question 13


Test: Ellipse- 3 - Question 14

The tangent at the point ‘a’ 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- 3 - Question 14


Equation of auxillary circle is x2 + y2 = a2
Let P is (acosa,bsina)

Equatio of AB is 

bcosα.x+a sinαy = ab
To get combined equation of CA and CB, homogenize equation of circle with equation (i),

b2x2 + b2y2 - (bcosα.x+a sinα.y)2 = 0
since ∠BCA = 90°
∴ coefficient of x2 + coefficient of y2 = 0


Test: Ellipse- 3 - Question 15

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

Detailed Solution for Test: Ellipse- 3 - Question 15

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

Test: Ellipse- 3 - 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- 3 - Question 16

For any tangent to ellipse


Using midpoint formula, we have



Test: Ellipse- 3 - Question 17

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

Test: Ellipse- 3 - 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 PQRS, is λ then 

Detailed Solution for Test: Ellipse- 3 - Question 18

Area of quadrilateral 

Test: Ellipse- 3 - 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- 3 - 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- 3 - 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- 3 - Question 21

Tangent at Q is x cos θ + y sin θ = a
ST = |a e cosθ - a| = a(1 - e cos θ)
Also SP = e PM
 
ST = SP ⇒ isosceles.

Test: Ellipse- 3 - 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- 3 - Question 22

Given equation is

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

Test: Ellipse- 3 - 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- 3 - 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- 3 - 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- 3 - 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- 3 - 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- 3 - 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- 3 - Question 26

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

Test: Ellipse- 3 - 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- 3 - 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- 3 - 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- 3 - 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- 3 - Question 29

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


Test: Ellipse- 3 - 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- 3 - Question 30

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


∴ b + c = 2s – a = 10.
Locus of A is an ellipse with major axis of length 10

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