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PQ is a chord ofthe parabola y^2=4x who...

PQ is a chord ofthe parabola `y^2=4x` whose perpendicular bisector meets the axis at M and the ordinate of the midpoint PQ meets the axis at N. Then the length MN is equal to

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The correct Answer is:
2
`P-=(t_(1)^(2),2t_(1))andQ-=(t_(2)^(2),2t_(2))`
L is the midpoint of PQ.
`:." "L-=((1)/(2)(t_(1)^(2)+t_(2)^(2)),(t_(1)+t^(2)))`
`"Slope of PQ"=(2)/(t_(1)+t_(2))`
`y-(t_(1)+t_(2))=(-(t_(1)+t_(2)))/(2)(x-(t_(1)^(2)+t_(2)^(2))/(2))`
Putting y = 0, we get
`M-=(2+(t_(1)^(2)+t_(2)^(2))/(2),0)`
`N-=((t_(1)^(2)+t_(2)^(2))/(2),0)`
`:." "MN=OM-ON=2`
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