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Parametric equations

Find the points of intersection of the parabola x = t^2 , y = 2t with the circle x^2 + y^2 -9x +4 = 0

I subsituted then factorised the equation, however the roots that I got from factorising was wrong. I have no idea why.

Thanks
Reply 1
Avatar for dvs
dvs
10
t^4 + 4t^2 - 9t^2 + 4 = 0
t^4 - 5t^2 + 4 = 0
(t^2 - 1)(t^2 - 4) = 0
t^2 = 1 or t^2 = 4
t = ±1 or t = ±2

=> x = 1 or x = 4
=> y = 2, -2 or y = 4, -4

Hence the points of intersection are:
(1, 2), (1, -2), (4, 4), (4, -4)
Reply 2
Avatar for DOJO
DOJO
OP
dvs
t^4 + 4t^2 - 9t^2 + 4 = 0
t^4 - 5t^2 + 4 = 0
(t^2 - 1)(t^2 - 4) = 0
t^2 = 1 or t^2 = 4
t = ±1 or t = ±2

=> x = 1 or x = 4
=> y = 2, -2 or y = 4, -4

Hence the points of intersection are:
(1, 2), (1, -2), (4, 4), (4, -4)


I have just realised what I did wrong :rolleyes:

After I factorised to get

(t^2 - 1)(t^2 - 4) = 0
t^2 = 1 or t^2 = 4
t = ±1 or t = ±2

I had forgotten to multiply 2 by ^2 as we split t^4 to t^2 for the quadratic, which as you rightly stated would have got me the values I wanted , -4 and 4, to subs into the corresponding eqn. Thank you so much.

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