Description

This mock test of Test: Conic Sections (CBSE Level)- 2 for JEE helps you for every JEE entrance exam.
This contains 25 Multiple Choice Questions for JEE Test: Conic Sections (CBSE Level)- 2 (mcq) to study with solutions a complete question bank.
The solved questions answers in this Test: Conic Sections (CBSE Level)- 2 quiz give you a good mix of easy questions and tough questions. JEE
students definitely take this Test: Conic Sections (CBSE Level)- 2 exercise for a better result in the exam. You can find other Test: Conic Sections (CBSE Level)- 2 extra questions,
long questions & short questions for JEE on EduRev as well by searching above.

QUESTION: 1

The radius of the circle passing through the foci of the ellipse and having its centre at (0, 3) is

Solution:

QUESTION: 2

The line y = c is a tangent to the parabola 7/2 if c is equal to

Solution:

y = x is tangent to the parabola

y=ax^{2}+c

if a= then c=?

y′ =2ax

y’ = 2(7/2)x =1

x = 1/7

1/7 = 2(1/7)^{2} + c

c = 1/7 * 2/49

c = 7/2

QUESTION: 3

The equation 2x^{2}+3y^{2}−8x−18y+35 = λ Represents

Solution:

Given the equation is,

2x^{2}+3y^{2}−8x−18y+35=K

Or, 2{x^{2}−4x+4} + 3{y^{2}−6y+9}=K

Or, 2(x−2)^{2} + 3(y−3)^{2} =K.

From the above equation it is clear that if K>0 then the given equation will represent an ellipse and for K<0, no geometrical interpretation.

Also if K=0 then the given equation will be reduced to a point and the point will be (2,3).

QUESTION: 4

The locus of a variable point whose distance from the point (2, 0) is 2/3 times its distance from the line x = 9/2 is

Solution:

QUESTION: 5

The axis of the parabola 9y^{2}−16x−12y−57 = 0 is

Solution:

QUESTION: 6

A and B are two distinct points, Locus of a point P satisfying |PA| + |PB| = 2k, a constant is

Solution:

QUESTION: 7

The eccentricity of the hyperbola x^{2}−y^{2} = 9 is

Solution:

QUESTION: 8

Locus of the point of intersection of the lines x = sec θ + tan θ and y = sec θ – tan θ is

Solution:

QUESTION: 9

The line y = m x + c, touches the parabola y^{2} = 4ax if

Solution:

QUESTION: 10

The equations x = at^{2}, y = 4at ; t ∈ R represent

Solution:

QUESTION: 11

t ∈ R represents

Solution:

P(x,y) = [(e^{t }+ e^{-t})/2 , (e^{t} - e^{-t})/2]

(e^{t} + e^{-t})/2 = x --------------------------(1)

(e^{t }- e^{-t})/2 = y --------------------------(2)

Adding (1) & (2)

2e^{t }= 2x + 2y

e^{t} = x + y

Eq (1) e^{t} + e^{-t} = 2x

e^{t} + 1/e^{t} = 2x

(e^{t})^{2} + 1 = 2x*e^{t}

(x+y)^{2} + 1 = 2x(x+y)

x^{2} + y^{2} + 2xy + 1 = 2x^{2} + 2xy

x^{2} + y^{2} + 1 = 2x^{2}

(x^{2})/(1)^{2} - (y^{2})/(1)^{2} = 1 {which represents hyperbola equation}

QUESTION: 12

The vertex of the parabola y^{2} = 4a(x−a) is

Solution:

QUESTION: 13

The two parabolas x^{2} = 4y and y^{2} = 4x meet in two distinct points. One of these is the origin and the other is

Solution:

QUESTION: 14

The equation of the directrix of the parabola x^{2} = −4ay is

Solution:

QUESTION: 15

The eccentricity ‘e’ of a parabola is

Solution:

QUESTION: 16

The ellipse

Solution:

QUESTION: 17

The equations x = a cos θ , y = b sin θ, 0 ≤ θ < 2π , a ≠ b, represent

Solution:

QUESTION: 18

The graph of the function f(x) i/x i.e. the curve y = 1/x is

Solution:

QUESTION: 19

The line y = c touches the parabola y^{2} = 4ax when

Solution:

QUESTION: 20

The parabolas x^{2} = 4y and y^{2} = 4x intersect

Solution:

QUESTION: 21

The lngth of the common chord of the parabolas y^{2} = x and x^{2} = y is

Solution:

QUESTION: 22

The number of points on X-axis which are at a distance c units (c < 3) from (2, 3) is

Solution:

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+x^{2}+9=c

⇒x^{2}−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.

QUESTION: 23

The eccentricity of the conic 9x^{2} − 16y^{2} = 144 is

Solution:

QUESTION: 24

The eccentricity of 3x^{2}+4y^{2} = 24 is

Solution:

QUESTION: 25

The angle between the tangents drawn from the origin to the circle = (x−7)^{2}+(y+1)^{2} = 25 is

Solution:

Let the equation of tangent drawn from (0,0) to the circle be y=mx. Then, p = a ⇒ 7m+1/(m^{2}+1)^{1/2}= 5

⇒24m^{2} + 14m−24=0

⇒12m^{2} + 7m−12=0

⇒m1m^{2} = −12/12 =−1

∴ Required angle = π/2

### Practice Paper 2 - Conic Sections

Doc | 1 Page

### Practice Test: Conic Sections, Class 11, Mathematics

Doc | 3 Pages

### Notes : Conic Sections

Doc | 3 Pages

### NCERT Solutions: Conic Sections

Doc | 33 Pages

- Test: Conic Sections (CBSE Level)- 2
Test | 25 questions | 25 min

- Test: Conic Sections (CBSE Level)- 1
Test | 25 questions | 25 min

- Test: Probability (CBSE Level)- 2
Test | 25 questions | 25 min

- Test: Probability (CBSE Level)- 2
Test | 25 questions | 25 min

- Test: Linear Inequalities (CBSE Level) - 2
Test | 25 questions | 25 min