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Question

Through the vertex O of the parabola y2=4ax chords OP,OQ are drawn at right angles to each other. If the locus of mid points of the chord PQ is a parabola, then its vertex is

A
(4a,0)
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B
(2a,0)
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C
(a,0)
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D
(3a,0)
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Solution

The correct option is C (4a,0)
Since the locus of the mid point of the chord of PQ is the parabola, hence its vertex is the point where the chord PQ meets the axis of the parabola.
Let the point P be (at2,2at)
Since the vertex of the original (given) parabola is (0,0), therefore the equation of OP is
y0x0=y2atxat2
y=2tx ...(i)
Since OQ is perpendicular to OP and it two passes through the origin, therefore, equation of OQ is
y=t2x ...(ii)
Substituting the y=t2x into the equation of the parabola, we can determine the intersection point of OQ and the parabola.
Therefore t2x24=4ax
xt2=16a
x=16at2
Hence y=8at
Therefore the equation of the PQ wil be
y+8atx16at2=y2atxat2
Since the axis of the parabola is the positive x axis, hence the vertex of the new parabola will be of the form (x,0)
Substituting y=0 in the above equation, we get
8at(xat2)=(2at)(x16at2)
8ax8a2t2=2at2x+32a2
8x8at2=32a2t2x
x(8+2t2)=32a+8at2
x(8+2t2)=4a(8+2t2)
Therefore x=4a
Hence the vertex is (4a,0).

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