The normal chord at a point `t` on the parabola y^2=4axsubtends a right angle at the vertex then t^2 is equal to what?

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)