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If the tangents at the points `Pa n dQ` on the parabola `y^2=4a x` meet at `T ,a n dS` is its focus, the prove that `S P ,S T ,a n dS Q` are in GP.

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The correct Answer is:
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Tangents to parabola `y^(2)=4axP(at_(1)^(2),2at_(1))andQ(at_(2)^(2),2at_(2))` meet at `T(at_(1)t_(2),a(t_(1)+t_(2)))`.
Now, `SP=a+at_(1)^(2)andSQ=a+at_(2)^(2)`
`:." "(ST)^(2)=(at_(1)t_(2)-a)^(2)+(a(t_(1)+t_(2)))^(2)`
`=a^(2)(1+t_(1)^(2))(1+t_(2)^(2))`
`=SPxxSQ`
Hence, SP, ST and SQ are in G.P.
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