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Question Number 23104 by lizan 123 last updated on 26/Oct/17
show tht the curve with parametric equcations  x=t^2  ,y=t^3  −9t intersect at the point (9,0).
$${show}\:{tht}\:{the}\:{curve}\:{with}\:{parametric}\:{equcations} \\ $$$${x}={t}^{\mathrm{2}} \:,{y}={t}^{\mathrm{3}} \:−\mathrm{9}{t}\:{intersect}\:{at}\:{the}\:{point}\:\left(\mathrm{9},\mathrm{0}\right). \\ $$
Answered by ajfour last updated on 26/Oct/17
y=t(t^2 −9)   ⇒  y^2 =x(x−9)^2   double root at x=9 , so  curve intersects itself at x=9 .
$${y}={t}\left({t}^{\mathrm{2}} −\mathrm{9}\right)\: \\ $$$$\Rightarrow\:\:{y}^{\mathrm{2}} ={x}\left({x}−\mathrm{9}\right)^{\mathrm{2}} \\ $$$${double}\:{root}\:{at}\:{x}=\mathrm{9}\:,\:{so} \\ $$$${curve}\:{intersects}\:{itself}\:{at}\:{x}=\mathrm{9}\:. \\ $$

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