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Q.

The length of chord of parabola  x2=4y having the equation, x2y+42=0  is

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a

63

b

32

c

211

d

82

answer is A.

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Detailed Solution

 x2=4y,x2y+42=0
Eliminate(y)  2x24x162=0
x1+x2=22,x1x2=16
Similarly eliminate(x),  2y220y+32=0,y1+y2=10,y1y2=16
Length  AB=(x1x2)2+(y1y2)2
 

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