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Test: 35 Year JEE Previous Year Questions: Conic Sections


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Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 1

Two common tangents to the circle x2 + y2 = 2a2 and parabola y2 = 8ax are [2002]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 1

Any tangent to the parabola y2 = 8ax is...(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).

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 2

The  normal at the point   (bt12 , 2bt1)  on a parabola meets the parabola again in the point (bt22, 2bt2) , then            [2003]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 2

Equation of the normal to a parabola y2 = 4bx at point
(bt12,2bt1) is y = – t1 x + 2bt1+ bt13

As given, it also passes through (bt22 ,2bt2) then

 

 

 ⇒ t2 + t1 =

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 3

The foci of the ellipse  and the hyper bola  coincide. Then the value of b2 is    [2003]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 3

∴ Foci = ( ±3 , 0)

∴ foci of ellipse = foci of hyperbola

∴ for ellipse ae = 3 but a = 4,

∴  

Then b2 = a2 (1-e2) ⇒ b=

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 4

If a ≠ 0 an d the lin e 2bx + 3cy + 4d=0 passes through the points of intersection of the parabolas y= 4ax and x= 4ay, th en [2004]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 4

Solving equations of parabolas y2 = 4ax and x= 4ay
we get (0, 0) and ( 4a, 4a)
Substituting in the given equation of line 2bx + 3cy + 4d = 0,
we get d = 0 and 2b +3c = 0
⇒ d2 + (2b + 3c)= 0

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 5

The eccentricity of an ellipse, with its centre at the origin, is. If one of the directrices is x = 4 , then the equation ofthe ellipse is: [2004]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 5

  e =   Directrix ,

Equation of elhipe is  3x2 + 4y2 = 12

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 6

Let P be the point ( 1, 0 ) and Q a point on the locus y2 = 8x .The locus of mid point of PQ is [2005]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 6

P = (1, 0)  Q = (h, k)

Such that K2 = 8h Let (α, β) be the midpoint of  PQ

2α - 1 = h 2β = k.
(2β) 2 = 8(2α - 1)

⇒ β2 = 4α - 2

⇒ y2 -4x+2=0.

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 7

The locus of a point P (α, β) moving under the condition th at the line y = αx + β is a tangen t to the hyper bola is [2005]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 7

Tangent to the hyperbola is

Given that y = αx + β is the tangent of hyperbola

⇒ m = α and a2 m2 -b22

∴ a2α2 -b22

Locus is a2 x2 - y= b2 which is hyperbola.

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 8

An ellipse has OB as semi minor axis, F and F ' its focii and the angle FBF ' is a right angle. Then the eccentricity of the ellipse is [2005]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 8

⇒ 2(a2e2 + b2)=4a2e2 ⇒

Also e2 = 1- b2 / a2 =1-e2

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 9

The locus of the vertices of the family of parabol as     [2006]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 9

Given parabola is  

∴ Vertex of parabola is 

To find locus of this vertex,

 64xy = 105

which is the required locus.

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 10

In an ellipse, the distance between its foci is 6 and minor axis is 8.  Then its eccentricity is [2006]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 10

2ae =6 ⇒ ae = 3 ; 2b =8 ⇒ b = 4

b2 = a2(1-e2) ;16 = a2- a2e2 ⇒ a2 = 16 + 9= 25

⇒ a = 5 

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 11

Angle between the tangents to the curve y = x2 - 5x+6 at the points (2, 0) and (3, 0) is [2006]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 11

     ∴ m1 = (2x - 5)(2, 0) =-1 ,

m2 = (2x - 5)(3, 0)=1 ⇒ m1m2 =-1

i.e. the tangents are perpendicular to each other.

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 12

For the Hyper bola which of the following remains constant when a varies = ? [2007]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 12

Given, equation of hyperbola is

We know that the equation of hyperbola is

Here, a2 = cos2α and b2 = sin2α

We know that, b2 = a2 (e- 1)

⇒ sin2α = cos2α(e- 1)

⇒ sin2α + cos2α = cos2α.e2

⇒ e2 = 1 + tan2α  = sec2α  ⇒ e = secα 

Co-ordinates of foci are (± ae, 0) i.e. ( ± 1, 0)

Hence, abscissae of foci remain constant when α varies.

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 13

The equation of a tangent to the parabola y2 = 8x is y = x + 2. The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is [2007]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 13

Parabola y2 = 8x Y

We know that the locus of point of intersection of two perpendicular tangents to a parabola is its directrix.
Point must be on the directrix of parabola

∵ equation of directrix  x + 2 = 0 ⇒ x = –2

Hence the point is (–2, 0)

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 14

The normal to a curve at P(x, y) meets the x-axis at G. If the distance of G from the origin is twice the abscissa of P, then the curve is a [2007]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 14

Equation of normal at P(x, y) is Y – y  = 

Coordinate of G at X axis is (X, 0) (let)

⇒ 

∴ Co-ordinate of G

Given distance of G from origin = twice of the abscissa of P.
∵ distance cannot be –ve, therefore abscissa x should be +ve

On Integrating ⇒  ⇒ x2 – y2 = –2c1

∴ the curve is a hyperbola

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 15

A focus of an ellipse is at the origin. The directrix is the line x = 4 and the eccentricity is . Then the length of the semi-major axis is [2008]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 15

Perpendicular distance of directrix from focus

∴ Semi major axis = 8/3

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 16

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]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 16

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)

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 17

The ellipse  x2 + 4y= 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]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 17

The given ellipse is 

So A = (2, 0) and B= (0, 1) If PQRS is the rectangle in which it is inscribed, then P = (2, 1).

 be the ellipse circumscribing the rectangle PQRS.

Then it passes through P (2,1 )

Also, given that, it passes through (4, 0)

⇒ b2 = 4/3 [substituting a2 = 16 in eqn (a)]

∴ The required ellipse is   or   x2 + 12y2 =16

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 18

If two tangents drawn from a point P to the parabola y2 = 4x are at right angles, then the locus of P is [2010]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 18

The locus of perpendicular tangents is directrix i.e., x =-1

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 19

Equation of the ellipse wh ose axes ar e th e axes of coordinates and which passes through the point (–3, 1) and has eccentricity  is              [2011]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 19

Let the ellipse be

It passes through (– 3, 1) so

Also, b2 =a2 (1- 2 / 5) ⇒ 5b2 =3a2 ...(ii)

Solving (i) and (ii) we get a2  

So, the equation of the ellipse is 3x2 + 5y= 32

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 20

Statement-1 : An equation of a common tangent to the parabola y2 = 16x and the ellipse 2x2 + y=4 is y = 2x+ 2
Statement-2 : If the line  is a common tan gent to the parabola y2 = 16x and the ellipse 2x2 + y2 = 4, then m satisfies m4 + 2m2 = 24    [2012] [2012]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 20

Given equation of ellipse is 2x2 + y2 = 4

Equation of tangent to the ellipse 

...(1)

(∵ equation of tangent to the ellipse

is y = mx + c where

Now, Equation of tangent to the parabola

...(2)

(∵ equation of tangent to the parabola y2 = 4ax is

On comparing (1) and (2), we get

Squaring on both the sides, we get 16 (3) = (2m2 + 4) m2
⇒   48 = m2 (2m2 + 4) ⇒    2m4 + 4m2 – 48 = 0 ⇒ m4 + 2m2 –24 = 0
⇒    (m2 + 6)(m2 – 4) = 0 ⇒ m2= 4 (∵ m2 ≠ – 6) ⇒ m = ± 2
⇒ Equation of common tangents are
Thus, statement-1 is true.
Statement-2 is obviously true.

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 21

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]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 21

Equation of circle is (x – 1)2 + y2 =1 ⇒ radius = 1 and diameter = 2
∴ Length of semi-minor axis is 2.
Equation of circle is x2 + (y – 2)2 = 4 = (2)2 ⇒ radius = 2 and diameter = 4
∴ Length of semi major axis is 4 We know, equation of ellipse is given by

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 22

The equation of the circle passing through the foci of the ellipse   and having centre at (0, 3) is [JEE M 2013]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 22

From the given equation of ellipse, we have

a = 4, b = 3,

Now, radius of this circle =

⇒ Focii = (±, 0)

Now equation of circle is (x – 0)2 + (y – 3)2 = 16
x2 + y2 – 6y – 7 = 0

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 23

Given : A circle, 2x2 + 2y2 = 5 and a parabola, 
Statement-1 : An equation of a common tangent to these curves is 
Statement-2 : If the line,  is theircommon tangent, then m satisfies m4 – 3m2 + 2 = 0. [JEE M 2013] 

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 23

Let common tangent be

Since, perpendicular distance from centre of the circle to the common tangent is equal to radius of the circle,

therefore  

On squaring both the side, we get m2 (1 + m2) = 2
⇒ m4 + m2 – 2 = 0  
⇒ (m2 + 2)(m2 – 1) = 0
⇒ m = ± 1

both statements are correct as m = ±1 satisfies the given equation of statement-2.

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 24

 The locus of the foot of perpendicular drawn from the centre of the ellipse x2 + 3y2 = 6 on any tangent to it is [JEE M 2014]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 24

Given eqn of  ellipse can be written as

 a2 = 6, b2 = 2

Now, equation of any variable tangent is

where m is slope of the tangent So, equation of perpendicular line drawn from centre to tangent is

...(ii)
Eliminating m, we get

(x4 + y4 + 2x2 y2) = a2x2+ b2y2

⇒ ( x2 + y2)2 = a2 x+ b2y2

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 25

The slope of the line touching both the parabolas y2 = 4x and x2 =-32y is [JEE M 2014]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 25

Let tangent to y2 = 4x be y =

Since this is also tangent to x2 = – 32y

Now, D = 0

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 26

Let O be the vertex and Q be any point on the parabola, x2 = 8y. If the point P divides the line segment OQ internally in the ratio 1 : 3, then locus of P is : [JEE M 2015]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 26

Let P(h, k) divides OQ in the ratio 1 : 3
Let any point Q on x2 = 8y is (4t, 2t2).

Then by section formula

⇒ 2k =h2

Required locus of P is x2 = 2y

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 27

The normal to the curve, x2 + 2xy – 3y2 = 0, at (1, 1) [JEE M 2015]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 27

Given curve is x2 + 2xy – 3y2 = 0    ...(1)

Differentiate w.r.t. x, 

Equation of normal at (1, 1) is y = 2 – x ...(2)
Solving eq. (1) and (2), we get x = 1 , 3
Point of intersection (1, 1), (3, –1)
Normal cuts the curve again in 4th quadrant.

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 28

The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse     is : [JEE M 2015]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 28

The end point of latus rectum of ellipse

in first quadrant is   and the tangent at this point intersects x-axis at and y-axis at (0, a).

The given ellipse is 

Then a2 = 9, b2 = 5   ⇒

∴ end point of latus rectum in first quadrent is L (2, 5/3)
Equation of tangent at L is 

It meets x-axis at A (9/2, 0) and y-axis at B (0, 3)

∴ Area of ΔOAB  

By symmetry area of quadrilateral

= 4 × (Area ΔOAB) = sq. units.

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 29

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]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 29

Minimum distance
⇒ perpendicular distance Eqn of normal at p(2t2, 4t) y = –tx + 4t + 2t3

It passes through C(0, –6)  ⇒ t3 + 2t + 3 = 0  ⇒ t  = – 1

Centre of new circle = P(2t2 , 4t) = P(2,-4)

Radius 

∴ Equation of the circle is

⇒ x2 + y2 – 4x + 8y + 12 = 0

Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 30

The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is : [JEE M 2016]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Conic Sections - Question 30

⇒ 4b2 = a2e2 ⇒ 4a2(e2 -1)= a2e2 ⇒ 3e2 = 4

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