Parametric equations

C1 curve.... x=t+1 y=2t^2
C2 curve.... x=2t+1 y=t^2+7
what's the point of intersection?

for the second equation replace t with something else (e.g. u).

x=t+1 y=2t^2
x=2u+1 y=u^2+7

Now solve for t and u such that both the x and y coordinates match:
t+1 = 2u+1 (x match)
2t^2 = u^2+7 (y match)

You can rearrange the x equation to get:
t = 2u
Then replace t with 2u in the second equation
2(2u)^2 = u^2 + 7
etc.
Note you will get two solutions

Thank You, Paula !

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Equating the corresponding x's and y's,

... x = t + 1 ... and ... x = 2t + 1

∴ t = x - 1 ∴ x = 2(x-1) + 1 ∴ x = 2x - 2 + 1 ∴ x = 1 ....... (1)
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... y = 2t² ... and ... y = t² + 7

∴ t = √(y) / √(2) ∴ y = ( y/2 ) + 7

∴ 2y = y + 14 ∴ y = 14 ..................................... (2)
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From (1) and (2),

the point of intersection is

( x, y ) ≡ ( 1, 14 ) ........................................… Ans.
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