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C4 parametric equation help




I haven't ventured to the bottom to have a look at the mark scheme (because that doesn't normally help at all), but i'm so confused about what i'm supposed to do part 1 and how i'm supposed to do it. I realise it's probably fairly straight forward, but i just can't get my head around it.


Thanks in advance :smile:
(edited 13 years ago)
Post the question, for those of us without integral logins.
Original post by CharleyChester
http://integralmaths.org/resources/file.php/40/papers/C4-A.pdf Question 5.


I haven't ventured to the bottom to have a look at the mark scheme (because that doesn't normally help at all), but i'm so confused about what i'm supposed to do part 1 and how i'm supposed to do it. I realise it's probably fairly straight forward, but i just can't get my head around it.


Thanks in advance :smile:


Would you be able to copy out the question? I can't see it as you need to login to integral maths :s-smilie:
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CharleyChester
OP
Original post by TimmonaPortella
Post the question, for those of us without integral logins.



Original post by jameswhughes
Would you be able to copy out the question? I can't see it as you need to login to integral maths :s-smilie:


Sorry! I forgot about the log-in thing. I screen captured the picture and put the link in the original post.
dxdt=2at\frac{dx}{dt}=2at
dydt=2a\frac{dy}{dt}=2a
dydx=1t\frac{dy}{dx}=\frac{1}{t}

For tangent
y=mx+cy=mx+c
where
m=dydxm= \frac{dy}{dx}

Use y=2apy=2ap and x=ap2x = ap^2 as given, and t=pt=p so dydx=1p\frac{dy}{dx}=\frac{1}{p}

so 2ap=ap+c2ap=ap+c therefore c=apc=ap

You now have a line, y=xp+apy= \frac{x}{p} + ap
and to find where it crosses the x axis, y=0y=0
so x=−ap2x =-ap^2

T=−ap2T=-ap^2

I'll do N later, go to go now!
Original post by CharleyChester



I haven't ventured to the bottom to have a look at the mark scheme (because that doesn't normally help at all), but i'm so confused about what i'm supposed to do part 1 and how i'm supposed to do it. I realise it's probably fairly straight forward, but i just can't get my head around it.


Thanks in advance :smile:


If the question was just

"A point P on the curve f(x) = (equation) has coordinates (x,y).
i) Find the coordinates of the point where the tangent meets the x-axis,
ii) Find the coordinates of the point where the normal meets the x-axis"

would you know how to go about it? It's the same method, but with parametric coordinates instead of cartesian. You should know how to find the gradient of a curve defined parametrically.
(edited 13 years ago)
For the normal, repeat what I did earlier, with m=−(dxdy)m=- (\frac{dx}{dy}) because the normal is perpendicular to the tangent, so the gradient is the negative inverse.

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