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-

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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