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Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced


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18 Questions MCQ Test Maths 35 Years JEE Main & Advanced Past year Papers | Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced

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*Multiple options can be correct
Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 1

The n umber of values of c such that the straight line y = 4x + c touches the curve (x2/4) + y2 = 1 is (1998 - 2 Marks)

Detailed Solution for Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 1

The given curve is  (an ellipse) and given line is y = 4x + c.
We know that y = mx + c touches the ellipse

Here

∴ two values of c exist

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 2

If P = (x, y), F1 = (3, 0), F2 = (–3, 0) and 16x2 + 25y2 = 400, then PF1 +PF2 equals          (1998 - 2 Marks)

Detailed Solution for Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 2

The ellipse can be written as,

Here a2 = 25, b2 = 16, but b2 = a2 (1 – e2)
⇒ 16/25 = 1– e2 ⇒ e2 =1 – 16/25 = 9/25  ⇒ e = 3/5
Foci of the ellipse are (+ ae, 0) = (± 3, 0), i.e., F1 and F2
∴ We have PF1 + PF2 = 2a = 10 for every point P on the ellipse.

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 3

On the ellipse 4 x2 + 9y=1 , the points at which the tangents are parallel to the line 8x = 9y are(1999 - 3 Marks)

Detailed Solution for Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 3

Let   be the tangent to 

where  

and pts of contact are  

or 

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 4

The equations of th e common tangents to the parabola y = x2 and y = – (x – 2)2 is/are (2006 - 5M, –1)

Detailed Solution for Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 4

if y = mx + c is tangent to y = x2 then x2 – mx – c = 0 has equal roots

⇒ m2 + 4c = 0 ⇒ c

is tangent to y = x2

∴ This is also tangent to  y = – (x – 2)2

= x2 + 4x – 4

⇒ x2 + (m – 4)x += 0

has equal roots⇒ m2 – 8m + 16 = – m2 + 16

⇒ m = 0, 4 ⇒ y = 0 or y = 4x – 4 are the tangents.

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 5

Let a hyperbola passes through the focus of the ellipse . The transverse and conjugate axes of this hyperbola coincide with the major and minor axes of the given ellipse, also the product of eccentricities of given ellipse and hyperbola is 1, then (2006 - 5M, –1)

Detailed Solution for Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 5

For the given ellipse

⇒ Eccentricity of hyperbola 

Let the hyperbola be  = 1 then

 = 1
 As it passes through focus of ellipse
i.e. (3, 0)

∴ we get A2 = 9 ⇒ B2 = 16

∴ Equation of hyper bola isfocus of hyperbola is (5, 0), vertex of hyperbola is (3, 0).

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 6

Let P(x1, y1) and Q(x2, y2), y1 < 0, y2 < 0, be the end points of the latus rectum of the ellipse x2 + 4y2 = 4. The equations of parabolas with latus rectum PQ are (2008)

Detailed Solution for Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 6

Given ellipse is  x2 + 4y2 = 4

a = 2, b = 1

As per question 

∴ PQ = 2

Now if PQ is the length of latus rectum to be found, then PQ = 4a =

 Also as PQ is horizontal, parabola with PQ as latus rectum can be upward parabola (with vertex at A) or down ward parabola (with vertex at A').
For upward parabola,

AR = a =  ∴ Coordinates of A =

So equation of upward parabola is given by

or  

For downward parabola A'R = a = 

∴ Coordinates of A' =  

So equation of downward parabola is given by

  or 

∴ the equation of required parabola is given by equation (1) or (2).

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 7

In a triangle ABC with fixed base BC, the vertex A moves such that

cosB + cosC = .

If a, b and c denote the lengths of the sides of the triangle opposite to the angles A, B and C, respectively, then

Detailed Solution for Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 7

In ΔABC , given that

But in a triangle sin 

Applying componendo and dividendo, we get

⇒ a + b + c  = 3a or  b + c = 2a
i.e.  AC+ AB = constant        
(∵ Base BC = a is given to be constant) ⇒ A moves on an ellipse.

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 8

The tangent PT and the normal PN to the parabola y2 = 4ax at a point P  on it meet its axis at points T  and N,  respectively.The locus of the centroid of the triangle PTN is a parabola whose(2009)

Detailed Solution for Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 8

Let P(at2 , 2at) be any point on the parabola y2 = 4ax.

Then tangent to parabola at P is  

which meets the axis of parabola i.e x-axis at T (– at2, 0).
Also normal to parabola at P is tx + y = 2at+ at3 which meets the axis of parabola at N (2a + at2 , 0)
Let G (x, y) be the centriod of Δ PTN , then

Eliminating t from above, we get the locus of centriod G as

which is a parabola with vertex  directrix as

latus r ectum as   and
focus  as (a, 0).

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 9

An ellipse intersects the hyperbola 2x2 – 2y2 = 1 orthogonally.The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then (2009)

Detailed Solution for Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 9

The given hyperbola is

...(1)

which is a rectangular hyperbola (i.e.  a = b)

Let the ellipse be 

Its eccentricity  = 

So, the equation of ellipse becomes x2 + 2 y= a2 ...(2)
Let  the hyperbola (1) and ellipse (2) intersect each other at P( x1 ,y1).
Then slope of hyperbola (1) at P is given by

and that of ellipse (2) at P is

As the two curves intersect orthogonally,

∴ m1m2 =-1

Also P ( x1,y1) lies on x2 -y=

...(ii)

Solving (i) and (ii), we get 

Also P ( x1,y1) lies on ellipse x2 + 2y= a2

 1 + 1 = a2 or a2 =2

∴ The required ellipse is x2 + 2y= 2 whose foci

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 10

Let A and B be two distinct points on the parabola y2 = 4x. If the axis of the parabola touches a circle of radius r having AB as its diameter, then the slope of the line joining A and B can be
(2010)

Detailed Solution for Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 10

Given parabola y2 = 4x

Let  A(t12 ,2t1 ) and B(t22 ,2t2) Then centre of circle drawn with AB as diameter is

As circle touches x-axis

∴ r = | t1 + t2| ⇒ t1 + t2 = ±r

Also slope of  AB 

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 11

Let the eccentricity of the hyperbol a abbereciprocal to that of the ellipse x2 + 4y2 = 4. If the hyperbola passes through a focus of the ellipse, then (2011)

Detailed Solution for Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 11

For x2 + 4y2 = 4        or  

As per question, 

focus of ellipse is (±, 0) As hyperbola passes through (±, 0)

∴ b = 1 and focus of hyperbola (± 2, 0)

∴ Equation of hyperbola is  

or     x2 – 3y2 = 3

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 12

Let L be a normal to the parabola y2 = 4x. If L passes through the point (9, 6), then L is given by (2011)

Detailed Solution for Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 12

The equation of normal to y2 = 4x is y = mx – 2m – m3 As it passes through (9, 6)
∴ 6 = 9m – 2m – m3 ⇒ m3 – 7m + 6 = 0 ⇒ (m – 1) (m2 + m – 6) = 0
⇒ (m – 1) (m + 3) (m – 2) = 0 ⇒ m = 1, 2, –3
∴ Normal is y = x – 3 or y = 2x – 12 or y = – 3x + 33
∴ a, b, d are the correct option.

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 13

Tangents are drawn to the hyperbola parallel to the straight line 2x – y = 1. The points of contact of the tangents on the hyperbola are (2012)

Detailed Solution for Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 13

KEY CONCEPT : If slope of tangent is m, then equations of tangents to hyperbola

= are

 with the points of contact

∵ Tangent to hyperbola = is parallel to 2x – y = 1, therefore slope of tangent = 2

∴ Points of contact are  

i.e.

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 14

Let P and Q be distinct points on the parabola y2 = 2x such that a circle with PQ as diameter passes through the vertex O of the parabola. If P lies in the first quadrant and the area of the triangle ΔOPQ is 3, then which of the following is (are) the coordinates of P?      (JEE Adv. 2015)

Detailed Solution for Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 14

Let point P in first quadrant, lying on parabola y2 = 2x

be Let Q be the point  Clearly a > 0.

∵ PQ is the diameter of circle through P, O, Q

∴ ∠POQ = 90°  = –1 ⇒ ab = –4

⇒ b is negative.
Also ar. ΔPOQ = 3

 (using ab = –4)

As a is positive and b is negative, we have a – b = 3 

(using ab = –4)

⇒ a2 – 3a + 4 = 0 ⇒ a2 – 2a – a + 4 = 0
⇒ (a – 2) (a – ) = 0  ⇒ a = 2 ,

∴ Point P can be  or 

i.e. (4, 2) or (1, )

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 15

Let E1 and E2 be two ellipses whose centers are at the origin.
The major axes of E1 and E2 lie along the x-axis and the y-axis, respectively. Let S be the circle x2 + (y – 1)2 = 2. The straight line x + y = 3 touches the curves S, E1 and E2 at P, Q and R respectively. Suppose that PQ = PR =
If e1 and e2 are the eccentricities of E1 and E2, respectively, then the correct expression(s) is (are)       (JEE Adv. 2015)

Detailed Solution for Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 15

Let E1 := 1 where a > b

and E2 := 1 where c < d

Also  S : x2 + (y – 1)2 = 2
Tangent at P(x1, y1) to S is x + y = 3
To find point of contact put x = 3 – y in S.
We get P(1, 2) Writing eqn of tangent in parametric form

 and 

 and 

eqn of tangent to E1 at Q is

which is identical to

⇒ a2 = 5 and b2 = 4 ⇒ 

eqn of tangent to E2 at R is

 identical to 

⇒ c2 = 1, d2 = 8 ⇒  

 

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 16

Consider the hyperbola H : x2 – y2 = 1 and a circle S with center N(x2, 0). Suppose that H and S touch each other at a point P(x1, y1) with x1 > 1 and y1 > 0. The common tangent to H and S at P intersects the x-axis at point M. If (l, m) is the centroid of the triangle PMN, then the correct expression(s) is(are)        (JEE Adv. 2015)

Detailed Solution for Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 16

H : x2 – y2 = 1  S : Circle with centre N(x2, 0) Common tangent to H and S at P(x1, y1) is

xx1 – yy1 = 1   ⇒

Also radius of circle S with centre N(x2, 0) through point of contact (x1, y1) is perpendicular to tangent

⇒ x1 = x2 – x1 or x2 = 2x1

M is the point of intersection of tangent at P and x-axis

∵  Centroid of ΔPMN is (ℓ, m)

and y1 = 3m

Using x2 = 2x1,

Also (x1, y1) lies on H,

 

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 17

The circle C1 : x2 + y2 = 3, with centre at O, intersects the parabola x2 = 2y at the point P in the first quadrant. Let the tangent to the circle C1, at P touches other two circles C2 and C3 at R2 and R3, respectively. Suppose C2 and C3 have equal radii 2 and centres Q2 and Q3, respectively. If Q2 and Q3 lie on the y–axis, then (JEE Adv. 2016)

Detailed Solution for Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 17

C1 : x2 + y2 = 3 ..(i)
parabola : x2  = 2y ...(ii)
Intersection point of (i) and (ii) in first quadrant y2 + 2y – 3 = 0 ⇒ y = 1 (∵y ≠ -3)
∴x =
P( ,1)
Equation of tangent to circle C1 at P is x + y-3 = 0
Let centre of circle C2 be (0, k); r = 2

 k = 9 or –3

∴ Q2 (0, 9), Q3 (0, –3)
(a)Q2 Q3 = 12
(b) R2R3 = length of transverse common tangent

(c) Area  × length of ⊥ from originto tangent

(d) Area  distance of P from

y–axis =

*Multiple options can be correct
Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 18

 Let P be the point on the parabola y2 = 4x wh ich is at the shortest distance from the center S of the circle x2 + y2 – 4x –16y + 64 = 0. Let Q be the point on the circle dividing the line segment SP internally. Then (JEE Adv. 2016)

Detailed Solution for Test: MCQs (One or More Correct Option): Conic Sections | JEE Advanced - Question 18

Let point P on parabola y2 = 4x be (t2, 2t)

∵ PS is shortest distance, therefore PS should be the normal to parabola.

Equation of normal to y2 = 4x at P (t2, 2t) is y – 2t = – t(x – t2)
It passes through S(2, 8)
∴ 8 – 2t = – t(2 – t2) ⇒ t3 = 8 or t = 2
∴ P(4, 4)
Also slope of tangent to circle at

Equation of normal at t = 2 is 2x + y = 12
Clearly x-intercept = 6
SP = and SQ = r = 2
∴ Q divides SP in the ratio SP : PQ
Hence a, c , d are the correct options.

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