A normal chord AB of a parabola y2−12x=0 subtends a right angle at the vertex of the parabola. If the point of intersection of the normals drawn at A and B is (p,q), then the value of p2q2 is
A
2
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B
1
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C
12
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D
14
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Solution
The correct option is A2
y2=12x;y2=4ax⇒a=3⇒A:(3t21,6t1),B:(3t22,6t2)
OA is perpendicular to OB ⇒mOA×mOB=−1⇒6t13t21×6t23t22=−1⇒t1t2=−4⋯(1)
If AB is normal at A and intersects the parabola at B, then t2=−t1−2t1⇒t1t2=−t21−2⇒t1=±√2[From (1)]⇒t2=∓2√2
B is the point of intersection of the two normals at A and B Coordinates of B is (24,∓12√2)=(p,q) ⇒p2q2=2
If AB is normal at B and intersects the parabola at A, then t1=−t2−2t2⇒t1t2=−t21−2⇒t2=±√2⇒t1=∓2√2
A is the point of intersection of the normals at A and B Coordinates of A is (24,∓12√2) ⇒p2q2=2