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
View Answer

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

Answer: d
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)
View Answer

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

Answer: d
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
View Answer

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

Answer: c
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
View Answer

Answer: a
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.
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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
View Answer

Answer: c
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
View Answer

Answer: c
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.
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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
View Answer

Answer: b
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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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter