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Question

# The length of the chord of the parabola x2=4y having equation x−√2y+4√2=0 is :

A
32
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B
211
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C
63
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D
82
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Solution

## The correct option is C 6√3Let (x1,y1) and (x2,y2) be the point of intersection of the chord and the parabola. x2=4y ⋯(1) x−√2y+4√2=0 ⋯(2) From equation (1) and (2) x−√2⋅x24+4√2=0 ⇒√2x2−4x−16√2=0 ∴x1+x2=2√2, x1x2=−16 So, x1−x2=6√2 Now from equation (2) x1−√2y1+4√2=0 ⋯(a) x2−√2y2+4√2=0 ⋯(b) Substract (a) from (b) x2−x1=√2(y2−y1) ⇒(x2−x1)2=2(y2−y1)2 So, AB=√(x2−x1)2+(y2−y1)2 =√(x2−x1)2+(x2−x1)22 =√32×|x2−x1| =√32×6√2 =6√3

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