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The tangents to the parabola `y^2=4a x`at the vertex `V`and any point `P`meet at `Q`. If `S`is the focus, then prove that `S PdotS Q ,`and `S V`are in GP. |
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Answer» Let the parabola be `y^(2)=4ax`. Q is the intersection of the line x=0 and the tangent at point `P(at^(2),2at),ty=x+at^(2)`. Solving these, we get Q = (0, at). Also, S=(a,0). Now, focal length `SP=a+at^(2)` `SQ^(2)=a^(2)+a^(2)t^(2)=a^(2)(t^(2)+1)` and SV = a `:." "SQ^(2)=SPxxSV` Therefore, SP,SQ and SV are in GP. |
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