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Question Number 23181 by lizan 123 last updated on 27/Oct/17

show that the curve with parametric equcations x=t^2 −3t+5, y=t^3 +t^2 −10t+9 intersect at the point (3,1).

$${show}\:{that}\:{the}\:{curve}\:{with}\:{parametric}\: \\ $$$${equcations}\:\:{x}={t}^{\mathrm{2}} \:−\mathrm{3}{t}+\mathrm{5}, \\ $$$${y}={t}^{\mathrm{3}} \:+{t}^{\mathrm{2}} \:−\mathrm{10}{t}+\mathrm{9}\:{intersect}\:{at}\:{the}\: \\ $$$${point}\:\left(\mathrm{3},\mathrm{1}\right). \\ $$

Commented by mrW1 last updated on 27/Oct/17

x=t^2 −3t+5=3 ⇒t^2 −3t=−2 ⇒t^2 =3t−2 ⇒t^3 =3t^2 −2t y=t^3 +t^2 −10t+9 =3t^2 −2t +t^2 −10t+9 =4t^2 −12t +9 =4(t^2 −3t)+9 =4×(−2)+9 =1 ⇒the curve passes through the point (3,1)

$${x}={t}^{\mathrm{2}} \:−\mathrm{3}{t}+\mathrm{5}=\mathrm{3} \\ $$$$\Rightarrow{t}^{\mathrm{2}} \:−\mathrm{3}{t}=−\mathrm{2} \\ $$$$\Rightarrow{t}^{\mathrm{2}} \:=\mathrm{3}{t}−\mathrm{2} \\ $$$$\Rightarrow{t}^{\mathrm{3}} \:=\mathrm{3}{t}^{\mathrm{2}} −\mathrm{2t} \\ $$$$ \\ $$$${y}={t}^{\mathrm{3}} \:+{t}^{\mathrm{2}} \:−\mathrm{10}{t}+\mathrm{9} \\ $$$$=\mathrm{3t}^{\mathrm{2}} −\mathrm{2t}\:+{t}^{\mathrm{2}} \:−\mathrm{10}{t}+\mathrm{9} \\ $$$$=\mathrm{4t}^{\mathrm{2}} −\mathrm{12t}\:+\mathrm{9} \\ $$$$=\mathrm{4}\left(\mathrm{t}^{\mathrm{2}} −\mathrm{3t}\right)+\mathrm{9} \\ $$$$=\mathrm{4}×\left(−\mathrm{2}\right)+\mathrm{9} \\ $$$$=\mathrm{1} \\ $$$$ \\ $$$$\Rightarrow\mathrm{the}\:\mathrm{curve}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{the}\:\mathrm{point}\:\left(\mathrm{3},\mathrm{1}\right) \\ $$

Commented by ajfour last updated on 27/Oct/17

It has to intersect itself at (3,1) . and not just pass through the point.

$${It}\:{has}\:{to}\:{intersect}\:{itself}\:{at}\:\left(\mathrm{3},\mathrm{1}\right)\:. \\ $$$${and}\:{not}\:{just}\:{pass}\:{through}\:{the} \\ $$$${point}. \\ $$

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