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← JEE Main 2016 · 10 Apr · Shift 1
MathematicsSingle correct+4 / -1Question 81 of 90

JEE Main 2016 · 10 Apr · Shift 1, Mathematics question 81

P and Q are two distinct points on the parabola, y2=4xy^2=4x, with parameters t and t1t_1 respectively. If the normal at P passes through Q, then the minimum value of t12t_1^2 is :

Choose one option
Answer and worked solution

Official answer: D

The normal at t meets the parabola again at t1=−t−2tt_1=-t-\frac2t, so t12=t2+4t2+4≥8t_1^2=t^2+\frac4{t^2}+4\ge8, with equality at t2=2t^2=2.