Conic section For IIT-Jee , NDA ,air force 31-45 (SET 3 )

conic section

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Question no. 31

The value of b does the line y= x+b touches the ellipse 9x² + 16y²= 144 is.

looks_one ∓5

looks_two ∓3

looks_3 ∓2

looks_4 ∓4

Option looks_one ∓5

Solution :
y = x + b is tangent on ellipse 9x² + 16y²= 144 (x²/16 + y²/9= 1)
Equation of tangent :

Question no. 32

The straight line lx + my +n =0 touches the ecllipse x²/a² + y²/b² =1 if

looks_one a²l²+b²m²= n²

looks_two a²n² + b²a²=m²

looks_3 a²m² + l²n²=m²

looks_4 a²b² + l²m²=n²

option looks_one a²l²+b²m²= n²

Solution :
lx + my +n =0 is the tangent of x²/a² + y²/b² =1
Equation of tangent :

Question no. 33

If the angle between the straight lines joining foci and the end of minor axis of the ellipse x²/a² + y²/b² =1 is 90^o, then its eccentricity

looks_one 1/2

looks_two 1/√3

looks_3 1/√2

looks_4 1/2√2

option looks_3 1/√2

Solution :

slope of prependicular lines = -1

Question no. 34

The two parabola y²= 4a ( x-b₁) and x²= 4a( y-b₂) always touch one another b₁ and b₂ being variable parameters. Their point of contact lies on the curve is :

looks_one xy=a²

looks_two xy=2a²

looks_3 xy=4a²

looks_4 xy=6a²

option looks_3 xy=4a²

Soluion :
Suppose these curve are meeting at (x,y) , there will be common tangent to both the curve. it means slope of line will be same.
Curve : y²= 4a ( x-b₁)
2ydy/dx = 4a ⇒ dy/dx = 2a/y ......[1]
Curve 2 : x²= 4a( y-b₂)
2x = 4ady/dx ⇒ dy/dx = x/2a .......[2]
from [1] and [2]
2a/y = x/2a
xy=4a²

Question no. 35

Through the vertex O of a parabola y²= 4x chords OP and OQ are drawn at right angles to one another. The locus of the mid point of PQ.

looks_one y² = 4(x+2)

looks_two y² = 2(x-4)

looks_3 x² = 4(y-2)

looks_4 x² = 2(y+2)

option looks_two y² = 2(x-4)

Solution :

x = (x1 + x2)/2 , y = (y1+y2)/2
y1.y2/x1.x2 = -1 [ multiply ofslope of prependicular lines are -1] .....[1]
y1²= 4(x1) .....[2]
(y2)²= 4(x2) .....[3]
from [1] , [2] , [3]
y1.y2 = -16 , x1.x2 = -16
2y = y1+y2
squaring both the side
4y² = y1² + y2² + 2.y1.y2
4y² = 4(x1) + 4(x2) - 16*2
y² = (x1+x2) - 8
y² = (2x) - 8

Question no. 36

The asymptotes of the hyperbola xy -3y -2x = 0 is

looks_one x-3 = 0

looks_two y-2 = 0

looks_3 x+3 = 0

looks_4 y+2 = 0

option looks_two y-2 = 0 , looks_one x-3 = 0

very soon 6

Question no. 37

The value of c does the line y = 2x + c touches the hyperbola 16x²-9y² = 144 is

looks_one ∓2√5

looks_two ∓ √5

looks_3 ∓ 3√5

looks_4 ∓ 4√5

option looks_one ∓2√5

Solution :
a² = 9
b = 16
m = 2
General equation of tangent :

Question no. 38

If e and e' be the eccentricities of a hyperbola and it's conjugate , then 1/e² + 1/e'² is

looks_one 2

looks_two 3

looks_3 4

looks_4 1

option looks_4 1

Solution :

Question no. 39

The condition straight line x/l + y/m = 1 touches the parabola y²= 4a(x+b) is

looks_one m²(l+b) +al=0

looks_two m(l+b)² +al=0

looks_3 m(lm+b) +al²=0

looks_4 m(l+b) + a²l²=0

option looks_3 m(lm+b) +al²=0

very soon 9

Question no. 40

The length of focal chord of parabola y2= 4ax makes an angle α with the x - axis is

looks_one 4acosec²α

looks_two 2acosec²α

looks_3 4acosec²α/2

looks_4 2acosec²α/2

option

very soon 10

Question no. 41

The equation of the parabola with its vertex at (3,2) and its focus at (5,2) is

looks_one y² -8x -4y +28=0

looks_two 2y² -x -4y + 28=0

looks_3 y² +8x +2y -28=0

looks_4 2y² +3x -4y +28=0

option y² -8x -4y +28=0

solution :
(y-Α)² = 4a(x-Β) where (Α ,Β ) are the point of vertex and a is the distance of focus from vertex
a= 5-3 = 2
(y-3)² = 4(2)(x-2)
y² -8x -4y +28=0

Question no. 42

A parabola whose vertex is (-11/8 , 3/2) and focus is ( -15/8 , 3/2 ) . Find its length of latus rectum

looks_one 3

looks_two 5

looks_3 2

looks_4 4

option looks_3 2

solution :
a = -15/8 +11/8 = -1/2
latus rectum = |4a| = |4*(-1/2)| = 2

Question no. 43

The condition straight line lx+ my +n = 0 touches the parabola y² = 4ax is

looks_one lm=an

looks_two ln=am²

looks_3 lm=an²

looks_4 l²n=a²m

option looks_two ln=am²

Solution :
y² = 4ax
equation of tangent : y = m'x + a/m'
lx+ my +n = 0
rearrange this equation in the form y = m'x + a/m'
y = (-l/m)x - n/m
m' = -l/m , a/m' = -n/m
put the value of m' in second equation
a(m/l) = n/m
ln=am²

Question no. 44

The equation of common tangent to the circle x² + y² = 8 and parabola y² = 16x is

looks_one y = 2x+4

looks_two y = x+4

looks_3 y = x-4

looks_4 y = -x-4

option looks_two y = x+4 , looks_4 y = -x-4

Soluion :
Equation of tangent to parabola : y = mx + a/m
y = mx + 4/m
Equation of tangent to circle :

equation : y = mx + 4/m
y = x+4 , y = -x-4

Question no. 45

The circle drawn on any focal chord of parabola as diameter touches the directrix is

looks_one x=-a

looks_two x=a

looks_3 x=0

looks_4 none of these

option looks_one x=-a

Solution :

diameter = 4a
radius =2a ,
x = -a is the directrix where circle touches it.

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