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Question

If the tangent at P on y2=4ax meets the tangent at the vertex in Q, and S is the focus of the parabola, then SQP=

A
π3
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B
π4
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C
π2
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D
2π3
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Solution

The correct option is C π2
We are given y2=4ax
Let any point on parabola beP(at2,2at)
Tangent atP:y=xt+at
ty=x+at2(1)
And tangent at vertex is x=0(2)
Solving (1) and (2), we get
ty=at2
y=at
So, coordinates of Q(0,at) and S=(a,0)
Now, slope of QP=2atatat20=1t=m (say)
Slope of QS=0ata0=t=m say
Now, mm=1t×t=1
Hence SQP=π2

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