If the normal to the parabola y2=4ax at the point (at2,2at) cuts the parabola again at (aT2,2aT), then
A
−2≤T≤2
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B
T∈(−∞,−8)∪(8,∞)
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C
T2<8
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D
T2≥8
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Solution
The correct option is DT2≥8 Eqution of the normal of the parabola y2=4ax at the point (at2,2at) is y+tx=2at+at3...(i) ∵ equation (i) cuts the parabola again at (aT2,2aT).
Then, 2aT+taT2=2at+at3 ⇒2a(T−t)=−at(T2−t2) ⇒t2+tT+2=0(∵t≠T) ∵t is real, ∴T2−4⋅1⋅2≥0⇒T2≥8