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Parametric equation help please

Mathematics - parametric equations?
A curve is defined by the parametric equations x = 2t^2, y = 4t
(i)By eliminating the parameter, find the cartesian equation of the curve.
(ii) Find the equation of the tangent to the curve at the point A with parameter t = 2
(iii) show that the tangent does not meet the curve again.
(iv) the normal of the curve at A cuts the curve again at B. Find the coordinates of B.



In the standard spirit of TSR, I'll post what I've done so far so you know where to help me.

i) x = 2t^2, y = 4t
t=y/4
x = (1/8)y^2
y= sqrt(8x)

ii) dy/dx = (dy/dt)/(dx/dt) = 4/4t. When t=2, dy/dx = 1/2.

Using y=mx+c,

y=0.5x+c

Where do I go from here please? I'm confused with the extra t.
Reply 1
Avatar for TenOfThem
TenOfThem
18
Original post by jumblehunter
Mathematics - parametric equations?
A curve is defined by the parametric equations x = 2t^2, y = 4t
(i)By eliminating the parameter, find the cartesian equation of the curve.
(ii) Find the equation of the tangent to the curve at the point A with parameter t = 2
(iii) show that the tangent does not meet the curve again.
(iv) the normal of the curve at A cuts the curve again at B. Find the coordinates of B.



In the standard spirit of TSR, I'll post what I've done so far so you know where to help me.

i) x = 2t^2, y = 4t
t=y/4
x = (1/8)y^2
y= sqrt(8x)

ii) dy/dx = (dy/dt)/(dx/dt) = 4/4t. When t=2, dy/dx = 1/2.

Using y=mx+c,

y=0.5x+c

Where do I go from here please? I'm confused with the extra t.



Put t=2 into x and y to find your point
Reply 2
Avatar for mathz
mathz
8
what values do you get for x and y at the value t=2?
Reply 3
Avatar for jumblehunter
jumblehunter
OP
8
Original post by TenOfThem
Put t=2 into x and y to find your point


so 8=4+c
c=4?
y=0.5x+4?


You sure thats right. Thats not what I've been told the answer is. http://uk.answers.yahoo.com/question/index?qid=20100430113754AAcHGIu
(edited 11 years ago)
Reply 4
Avatar for TenOfThem
TenOfThem
18
Original post by jumblehunter
so 8=4+c
c=4?
y=0.5x+4?


You sure thats right. Thats not what I've been told the answer is. http://uk.answers.yahoo.com/question/index?qid=20100430113754AAcHGIu


That answer is the same

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