Consider a parabola y2=4ax. If the normal to the parabola at the point (at2,2at) cuts the parabola again at (aT2,2aT), then
A
T2≥8
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B
T2≤6
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C
T∈(−∞,−8)∪(8,∞)
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D
T∈(−8,8)
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Solution
The correct option is AT2≥8 Equation of normal to the parabola y2=4ax at the point (at2,2at) is y+tx=2at+at3⋯(1) Equation (1) cuts the parabola again at (aT2,2aT) Then, 2aT+taT2=2at+at3 ⇒2a(T−t)=−at(T2−t2) ⇒2=−t(T+t)[∵t≠T] ⇒t2+Tt+2=0 Since, t is real D≥0 ⇒T2−4⋅2⋅1≥0 ⇒T2≥8