If tangent at P and at vertex of the parabola y2=4ax meets at Q, then the value of ∠SQP, where S is the focus, is
A
π2
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B
π
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C
3π2
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D
π6
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Solution
The correct option is Aπ2 Given : y2=4ax
Let any point on parabola be P(at2,2at)
Tangent at P ty=x+at2⋯(1)
And tangent at vertex is x=0⋯(2)
Solving (1) and (2), we get ⇒y=at
So, the coordinates are Q≡(0,at)andS=(a,0)
Now, slopes are mQP=2at−atat2−0=1tmQS=0−ata−0=−t⇒mQP×mQS=−1∴∠SQP=π2