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4
4 Let `(t^(2),2t)` be a point on the curve `y^(2)=4xandQh,k)` be it's image in x+y+4=0.
`:.(h-t^(2))/(1)=(k-2t)/(1)=-(2(t^(2)+2t+4))/(2)`
`rArrh=-(2t+4)andk=-(t^(2)+4)`
Now y=-5 intersect this locus.
`:.h=-5,sot=pm1`
Hence, h=-2,-6
So, point of intersection are A(-2,-5) and B(-6,-5).
Hence, AB=4.
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