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

1. Find the equation of directrix of parabola x^{2}=100y.

a) x=25

b) x=-25

c) y=-25

d) y=25

View Answer

Explanation: Comparing equation with x

^{2}=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 x^{2}=-100y.

a) x=25

b) x=-25

c) y=-25

d) y=25

View Answer

Explanation: Comparing equation with x

^{2}=-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 y^{2}=4ax.

a) (0, 4)

b) (0, 0)

c) (4, 0)

d) (0, -4)

View Answer

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 y^{2}=24x.

a) x=0

b) x=6

c) y=6

d) y=0

View Answer

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

^{2}=24x is y=0.

5. Find the equation of axis of the parabola x^{2}=24y.

a) x=0

b) x=6

c) y=6

d) y=0

View Answer

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

^{2}=24y is x=0.

6. Find the length of latus rectum of the parabola y^{2}=40x.

a) 4 units

b) 10 units

c) 40 units

d) 80 units

View Answer

Explanation: Comparing equation with y

^{2}=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) x^{2}+y^{2}+2xy-4x-16y+54=0

b) x^{2}+y^{2}+2xy+4x-16y+54=0

c) x^{2}+y^{2}+2xy+4x+16y+54=0

d) x^{2}+y^{2}+2xy-4x+16y+54=0

View Answer

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}

x

^{2}+y

^{2}+2xy-4x-16y+54=0.

8. If vertex is at (1, 2) and focus (2, 0) then find the equation of the parabola.

a) y^{2}-8x+4y+12=0

b) y^{2}-8x-4y-12=0

c) y^{2}-8x-4y+12=0

d) y^{2}+8x+4y+12=0

View Answer

Explanation: Since vertex is at (1, 2) and focus (2, 0) so parabola equation will be

(y-2)

^{2}=4*2(x-1)

y

^{2}-8x-4y+12=0.

9. Equation of parabola which is symmetric about x-axis with vertex (0, 0) and pass through (3, 6).

a) y^{2}=6x

b) x^{2}=12y

c) y^{2}=12x

d) x^{2}=6y

View Answer

Explanation: Equation of parabola which is symmetric about x-axis with vertex (0, 0) is y

^{2}=4ax

Since parabola pass through (3, 6) then 6

^{2}=4a*3 => a=3.

So, equation is y

^{2}=12x.

10. Equation of parabola which is symmetric about y-axis with vertex (0, 0) and pass through (6, 3).

a) y^{2}=6x

b) x^{2}=12y

c) y^{2}=12x

d) x^{2}=6y

View Answer

Explanation: Equation of parabola which is symmetric about y-axis with vertex (0, 0) is x

^{2}=4ay

Since parabola pass through (6, 3) then 6

^{2}=4a*3 => a=3.

So, equation is x

^{2}=12y.

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