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).


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

$s h o w t h a t t h e c u r v e w i t h p a r a m e t r i c$ $e q u c a t i o n s x = t^{2} - 3 t + 5,$ $y = t^{3} + t^{2} - 10 t + 9 i n t e r s e c t a t t h e$ $p o i n t (3, 1) .$
Commented by mrW1 last updated on 27/Oct/17

$x = t^{2} - 3 t + 5 = 3$ $\Rightarrow t^{2} - 3 t = - 2$ $\Rightarrow t^{2} = 3 t - 2$ $\Rightarrow t^{3} = 3 t^{2} - 2 t$ $y = t^{3} + t^{2} - 10 t + 9$ $= {3 t}^{2} - 2 t + t^{2} - 10 t + 9$ $= {4 t}^{2} - 12 t + 9$ $= 4 (t^{2} - 3 t) + 9$ $= 4 \times (- 2) + 9$ $= 1$ $\Rightarrow the curve passes through the point (3, 1)$
Commented by ajfour last updated on 27/Oct/17

$I t h a s t o i n t e r s e c t i t s e l f a t (3, 1) .$ $a n d n o t j u s t p a s s t h r o u g h t h e$ $p o i n t .$
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