Let AB is a normal chord of parabola y2=4x, with foot of normal as A(1,−2). If AB=p√q, where p is a natural number and q is a prime number, then pq is equal to
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Solution
y2=4x⇒a=1
Parameter of A:2×1×t1=−2 ⇒t1=−1
Parameter of B:t2=−t1−2t1=3 ∴B(at22,2at2)=B(9,6) AB=8√2 ⇒p=8,q=2 ⇒pq=4