# Mathematics Questions and Answers – Conic Sections – Parabola-2

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This set of Mathematics Quiz for Class 11 focuses on “Conic Sections – Parabola-2”.

1. Find the equation of directrix of parabola x2=100y.
a) x=25
b) x=-25
c) y=-25
d) y=25

Explanation: Comparing equation with x2=4ay.
4a=100 => a=25. Directrix is a line parallel to latus rectum in such a way that vertex is at middle of both.
Equation of directrix is y=-a => y=-25.

2. Find the equation of directrix of parabola x2=-100y.
a) x=25
b) x=-25
c) y=-25
d) y=25

Explanation: Comparing equation with x2=-4ay.
4a=100 => a=25. Directrix is a line parallel to latus rectum in such a way that vertex is at middle of both.
Equation of directrix is y=a => y=25.

3. Find the vertex of the parabola y2=4ax.
a) (0, 4)
b) (0, 0)
c) (4, 0)
d) (0, -4)

Explanation: Vertex is close end point of the parabola i.e. origin (0, 0) as it satisfies the equation (y-0)2=4a(x-0) also.

4. Find the equation of axis of the parabola y2=24x.
a) x=0
b) x=6
c) y=6
d) y=0

Explanation: Axis is the line passing through focus which divides the parabola symmetrically into two equal halves. Equation of axis of parabola y2=24x is y=0.

5. Find the equation of axis of the parabola x2=24y.
a) x=0
b) x=6
c) y=6
d) y=0

Explanation: Axis is the line passing through focus which divides the parabola symmetrically into two equal halves. Equation of axis of parabola x2=24y is x=0.

6. Find the length of latus rectum of the parabola y2=40x.
a) 4 units
b) 10 units
c) 40 units
d) 80 units

Explanation: Comparing equation with y2=4ax.
4a=40. Length of latus rectum of parabola is 4a =40.

7. If focus of parabola is F (2, 5) and equation of directrix is x + y=2, then find the equation of parabola.
a) x2+y2+2xy-4x-16y+54=0
b) x2+y2+2xy+4x-16y+54=0
c) x2+y2+2xy+4x+16y+54=0
d) x2+y2+2xy-4x+16y+54=0

Explanation: We know, if P is a point on parabola, M is foot of perpendicular drawn from point P to directrix of parabola and F is focus of parabola then PF=PM
(x-2)2+(y-5)2 = (|x+y-2|/√2)2
x2+y2+2xy-4x-16y+54=0.

8. If vertex is at (1, 2) and focus (2, 0) then find the equation of the parabola.
a) y2-8x+4y+12=0
b) y2-8x-4y-12=0
c) y2-8x-4y+12=0
d) y2+8x+4y+12=0

Explanation: Since vertex is at (1, 2) and focus (2, 0) so parabola equation will be
(y-2)2=4*2(x-1)
y2-8x-4y+12=0.

9. Equation of parabola which is symmetric about x-axis with vertex (0, 0) and pass through (3, 6).
a) y2=6x
b) x2=12y
c) y2=12x
d) x2=6y

Explanation: Equation of parabola which is symmetric about x-axis with vertex (0, 0) is y2=4ax
Since parabola pass through (3, 6) then 62=4a*3 => a=3.
So, equation is y2=12x.

10. Equation of parabola which is symmetric about y-axis with vertex (0, 0) and pass through (6, 3).
a) y2=6x
b) x2=12y
c) y2=12x
d) x2=6y

Explanation: Equation of parabola which is symmetric about y-axis with vertex (0, 0) is x2=4ay
Since parabola pass through (6, 3) then 62=4a*3 => a=3.
So, equation is x2=12y.

Sanfoundry Global Education & Learning Series – Mathematics – Class 11.

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