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Question

Through the vertex O of the parabola y2=4ax a perpendicular is drawn to any tangent meeting it at P and the parabola at Q. Then OPOQ=

A
a2
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B
2a2
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C
3a2
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D
4a2
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Solution

The correct option is D 4a2
Here O is the vertex,
Equation of tangent at (at2,2at) is yt=x+at2 (1)
Distance
OP=|00at2|1+t2OP=|at2|1+t2

Equation of perpendicular to line (1)from O(0,0) is
y+tx=0 (2)
Now, intersection of y2=4ax and line (2) is
Q(4at2,4at)
Distance
OQ=(4at2)2+(4at)2OQ=|4a|1t4+1t2OQ=|4a|1+t2t2
Now,
OPOQ=|at2|1+t2×|4a|1+t2t2OPOQ=4a2

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