12th Grade > Mathematics
INTRODUCTION TO CONICS AND PARABOLA MCQs
Total Questions : 14
| Page 2 of 2 pages
Answer: Option B. -> (15.25),(15,−25)
:
B
y2=4.15x;a=15. Focus is (15,0) and co-ordinates of latus rectum are y2=425⇒y=±25
or end points of latus rectum are (15,±25) .
:
B
y2=4.15x;a=15. Focus is (15,0) and co-ordinates of latus rectum are y2=425⇒y=±25
or end points of latus rectum are (15,±25) .
Answer: Option B. -> 8
:
B
Let the equation of parabola is x2=4ay, but a = 4−2=−2.Thenequationisx2=−8y and latus rectum = 4a = 8
:
B
Let the equation of parabola is x2=4ay, but a = 4−2=−2.Thenequationisx2=−8y and latus rectum = 4a = 8
Answer: Option A. -> (0, 0), (4, 12)
:
A
y1=3x1 . According to given condition 9x21=36x1
⇒x1=4,0⇒y1=12,0
Hence the points are (0,0) and (4,12).
:
A
y1=3x1 . According to given condition 9x21=36x1
⇒x1=4,0⇒y1=12,0
Hence the points are (0,0) and (4,12).
Answer: Option C. -> x2=−12y
:
C
Since the axis of parabola is y-axis
∴ Equation of parabola x2=4ay
Since it passes through (6 , -3)
∴ 36 = -12 a ⇒a=−3
∴ Equation of parabola is x2=−12y .
:
C
Since the axis of parabola is y-axis
∴ Equation of parabola x2=4ay
Since it passes through (6 , -3)
∴ 36 = -12 a ⇒a=−3
∴ Equation of parabola is x2=−12y .
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