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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 & the parabola at Q. Show that OPOQ=constant.

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Solution

We assume that O is vertex, P is point where perpendicular meets tangent. Equation of tangent y2=4ax at (at[2],2at)
yt=2 t at2
xyt+at2=0 ------- (1)
Length of perpendicular from O(0,0) is
OP=|O+O+at2|1+t2
OP=|at2|1+t2
Equation of perpendicular to ,i, is
tx+y=k
It passes through (O,O)
O+O=k
tx+y=O
y=tx put in y2=4ax
t2x2=4ax
x=4at2
y=tx=t4at2=4at
Hence Q(4at2,4at)
By distance formula ,QB=(4at20)2+(4at0)2=|4a|1t4+1t2
=|4a|1+t2t4.OQ=4at21+t2,OP.OQ=4a2.

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