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Dear Sunny
let end point of the chord is (at12 ,2at1) and (at22 ,2at2)
since it is normal to parabola so t2 = -t1 - 2/t1
equation of normal of parabola y =-tx + 2at + at3
slope is -t1 which is given acute so -t1 >0
or t1
now chord make right angel at origin so {(2at1 -0)/ (at12 -0)}{(2at1 -0)/ (at12 -0)} =-1
t1t2 =-4
t1(-t1 - 2/t1) =-4
t1 =-√2
Hence
t² = 2
Regards
Arun (askIITians forum expert)
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