Conic Section
Question no. 1
The points on the parabola y2 = 36x whose ordinate is three times the abscissa are
looks_one (0,0), (4,12)
looks_two (1,3) ,(4,12)
looks_3 (4,12)
looks_4 none of these
Option looks_one (0,0), (4,12)
Solution :
ordinate → y
abscissa → x
y = 3x
y2 = 36x ⇒ 9x2 = 36x
9x(x-4) = 0 ⇒ x =0 , 4
y = 0 , 12
points are (0,0) and (4,12)
ordinate → y
abscissa → x
y = 3x
y2 = 36x ⇒ 9x2 = 36x
9x(x-4) = 0 ⇒ x =0 , 4
y = 0 , 12
points are (0,0) and (4,12)
Question no. 2
The focal distance of a point on the parabola y2 = 16x whose ordinate is twice the abscissa , is
looks_one 6
looks_two 8
looks_3 10
looks_4 12
option looks_two 8
Solution :
y2= 4ax
y = 2x
y2 = 16x
4x2 = 16x ⇒ x= 4 , 0
y = 8 ,0
focal point of given parabola = (4,0)
distance between between (4,8) and (4,0) is 8
y2= 4ax
y = 2x
y2 = 16x
4x2 = 16x ⇒ x= 4 , 0
y = 8 ,0
focal point of given parabola = (4,0)
distance between between (4,8) and (4,0) is 8
Question no. 3
The focus of the parabola x2=-16y is
looks_one (4,0)
looks_two (0,4)
looks_3 (-4 ,0)
looks_4 (0,-4)
option looks_4 (0,-4)
Solution :

x2=-16y
x2=-4ay ⇒ focus =(0 , -a )
x2=-16y ⇒ a = 4
focus = (0 ,-4)

x2=-16y
x2=-4ay ⇒ focus =(0 , -a )
x2=-16y ⇒ a = 4
focus = (0 ,-4)
Question no. 4
If (2,0) is the vextex and y axis the directrix of a parabola ,then its focus is
looks_one (2,0)
looks_two (-2,0)
looks_3 (4,0)
looks_4 (-4,0)
option looks_3 (4,0)
Solution :

y axis is the directrix of given parabola , it means given parabola is symmetric along x axis
directrix distance from the vertex is equal to the distance of focus from the vertex
directrix is 2 unit from vertex , so focus will be 2 unit from vertex
vertex point = (2 ,0) ⇒ focus point = (4 ,0)

y axis is the directrix of given parabola , it means given parabola is symmetric along x axis
directrix distance from the vertex is equal to the distance of focus from the vertex
directrix is 2 unit from vertex , so focus will be 2 unit from vertex
vertex point = (2 ,0) ⇒ focus point = (4 ,0)
Question no. 5
The ends of latus rectum of parabola x2 + 8y =0
looks_one (-4 ,-2) and (4,2)
looks_two (4 ,-2) and (-4,2)
looks_3 (-4 ,-2) and (4,-2)
looks_4(-4 ,2) and (4,2)
option looks_3 (-4 ,-2) and (4,-2)
Solution :
x2 + 8y =0 ⇒ x2 = -8y
a = 2
focus point = (0,-2)
latus rectum ends point = (x1, -2) , (x2 , -2)
put the points in the equations
x2 = -8y ⇒ x2 = 16 ⇒ x = 4 ,-4
points of latus point = (-4 ,-2) and (4,-2)
x2 + 8y =0 ⇒ x2 = -8y
a = 2
focus point = (0,-2)
latus rectum ends point = (x1, -2) , (x2 , -2)
put the points in the equations
x2 = -8y ⇒ x2 = 16 ⇒ x = 4 ,-4
points of latus point = (-4 ,-2) and (4,-2)
Question no. 6
Foci and directrices of the parabola x2 = -8ay are
looks_one (0,-2a ) and y=2a
looks_two (0,-2a ) and y=-2a
looks_3 (2a,0 ) and x=-2a
looks_4 (-2a ,0 ) and x=2a
option looks_one (0,-2a ) and y=2a
Solution :

x2 = -8ay (x2 = -4Ay) ⇒ A = 2a
foci : (0 , -2a)
directrix : y = 2a

x2 = -8ay (x2 = -4Ay) ⇒ A = 2a
foci : (0 , -2a)
directrix : y = 2a
Question no. 7
The ends of latus rectum of parabola x2= 4ay.
looks_one (a ,2a), (2a ,-a)
looks_two (-a ,2a), (2a ,a)
looks_3 (a ,-2a), (2a ,a)
looks_4 (-2a ,a), (2a ,a)
option looks_4 (-2a ,a), (2a ,a)
Solution :
latus rectum = (-2a, a), (2a ,a)
latus rectum = (-2a, a), (2a ,a)
Question no. 8
The equation of the parabola with focus(3,0) and the directrix x+3=0
looks_one y2= 3x
looks_two y2= 2x
looks_3 y2= 12x
looks_4 y2= 6x
option looks_3 y2= 12x
Solution :

focus = (3,0)
directrix x+3 = 0 ⇒ a = 3
parabola is symmetrical to x axis
y2 = 4ax
put the value of a = 3
y2 = 4*3*x = 12x

focus = (3,0)
directrix x+3 = 0 ⇒ a = 3
parabola is symmetrical to x axis
y2 = 4ax
put the value of a = 3
y2 = 4*3*x = 12x
Question no. 9
The parabola y2= x is symmetric about
looks_one y-axis
looks_two x-axis
looks_3 both x-axis and y-axis
looks_4 the line y=x
option looks_two x-axis
Solution:

The parabola y2= x is symmetric about x-axis

The parabola y2= x is symmetric about x-axis
Question no. 10
Vertex of the parabola y2 + 2y +x= 0 lies in the quadrant
looks_one first
looks_two second
looks_3 third
looks_4 fourth
option looks_4 fourth
Solution :
y2 + 2y +x= 0 ⇒ (y+1)2 = -x + 1 = -(x-1)
y2 + 2y +x= 0 ⇒ (y+1)2 = -x + 1 = -(x-1)
Question no. 11
If the latus rectum of an ellipse is equal to half its minor axis, then what is its eccentricity?
looks_one 1/2
looks_two √3
looks_3 √3/2
looks_4 1/√2
option looks_3 √3/2
Solution :


Question no. 12
What are the points of intersection of the curve 4x2 - 9y2 = 1 with its conjugate axis?
looks_one (1/2 ,0) and (-1/2 ,0)
looks_two (0,2) and (0,-2)
looks_3 (0,3) and (0,-3)
looks_4 No such point exist
option looks_4 No such point exist
Solution :
conjugate axis never cut the hyperbola , therefore no such point exit
conjugate axis never cut the hyperbola , therefore no such point exit
Question no. 13
What is the sum of the local distances of a point of an ellipse x2/a2 + y2/b2 =1 ?
looks_one a
looks_two b
looks_3 2a
looks_4 2b
option looks_3 2a
Solution :

distance of f'A and fA = sum of local points of a ellipse
f'A = a + ae
fA = a-ae
f'A + fA = 2a

distance of f'A and fA = sum of local points of a ellipse
f'A = a + ae
fA = a-ae
f'A + fA = 2a
Question no. 14
What is the eccentricity of the conic 4x2 +9y2 = 144.
looks_one √5/3
looks_two √5/4
looks_3 3/√5
looks_4 2/3
option looks_one √5/3
Solution :


Question no. 15
The sum of the focal distances of a point on the ellipse x2/4 + y2/9 = 1 is
looks_one 4
looks_two 6
looks_3 8
looks_4 10
option looks_two 6
Solution :
sum of the focal distances = 2b = 2*3 = 6
sum of the focal distances = 2b = 2*3 = 6
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