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Solution
The correct option is A(2t[(t2+2)tan2α+1]1−t2tan2α,t2−(t2+2)2tan2α1−t2tan2α) Let any variable point on parabola be (λ2,2λ)
So, ∣∣
∣
∣
∣∣2λ+t−1t1+2t(λ+t)∣∣
∣
∣
∣∣=tanα ⇒λ2(1−t2tan2α)−2λt(1+(t2+2)tan2α)+t2−(t2+2)2tan2α=0
This equation in λ has two roots t1&t2
So, t1+t2=2t(1+(t2+2)tan2α)1−t2tan2α
and t1t2=t2−(t2−2)2tan2α1−t2tan2α (t1+t2)att=1=2(3tan2α+101−tan2α
Range will be (−∞,−6)∪[2,∞)