If the normal to y2=4ax at point t1 cuts the parabola again at point t2,then find the range of t2.
My method:
Pt. t1=(a(t1)2,2at1)
Pt. t2=(a(t2)2,2at2)
Equn. of normal at t1:y=-t1x + at1(t12 + 2)
Since the above equn. satisfies t2,we get:
2at2=-t1(a(t2)2) + at1(t12 + 2).
What next sir?
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2 Answers
62If the normal to y2=4ax at point t1 cuts the parabola again at point t2,then find the range of t2.
My method:
Pt. t1=(a(t1)2,2at1)
Pt. t2=(a(t2)2,2at2)
Equn. of normal at t1:y=-t1x + at1(t12 + 2)
Since the above equn. satisfies t2,we get:
2at2=-t1(a(t2)2) + at1(t12 + 2).
the above eqth is a quadratic in t
so one root is t1 the other is given by? (Sum of roots!)
hence you get both the roots :)
There are small issues in the final answer.. please look at them and fix everything accordingly
If you cant then please write in again!
1The 2 roots of t2 are: (t1) and (-(t12 + t + 2)).
Now, is: t22 =[2,8]
or
t22 = [0,2]
or
t22 = [8,∞]
or
None of the above?