x Turn on thread page Beta
 You are Here: Home >< Maths

FP1 - coordinate systems (question) watch

1. The point P(3t^2 , 6t) lies on the parabola C with equation y^2 = 12x.
(a) Show that an equation of the tangent to C at P is yt = x + 3t^2

y = sqrt(12) x sqrt(x)

dy/dx = (0.5)[sqrt(12)][x^(-1/2)]

dy/dx = sqrt(12)/2(sqrt)x

x = 3t^2, so...

dy/dx = sqrt(12)/2[sqrt(3t^2)]

dy/dx = sqrt(12)/2t[sqrt(3)]

y - 6t = [sqrt(12)/2t sqrt(3)] (x - 3t^2) .... (is this correct so far?)

multiplying both sides by 2t[sqrt(3)]...

(y-6t)(2t sqrt(3)) = [sqrt(12)] (x-3t^2)

simplifying...

2yt sqrt(3) - 12t^2 sqrt(3) = 2x sqrt(3) - 6t^2 sqrt(3) ... (i changed sqrt (12) to 2 sqrt(3))

simplifying further...

yt = 6t^2 - 6xt^2, since x = 3t^2...
yt = 2x - 2x^2

anyone know where i went wrong? would be MUCH appreciated!
2. You'd make your life easier if you wrote earlier on, because then the s disappear completely.

Everything up to is correct. From here, use my above advice, and the correct answer is about 2 lines away.

In fact everything is right up until your "simplifying further" bit. You can't substitute , since the 'x' here refers to the tangent line rather than the curve.
3. (Original post by nuodai)
You'd make your life easier if you wrote earlier on, because then the s disappear completely.

Everything up to is correct. From here, use my above advice, and the correct answer is about 2 lines away.

In fact everything is right up until your "simplifying further" bit. You can't substitute , since the 'x' here refers to the tangent line rather than the curve.
THANK YOU!
4. (Original post by nuodai)
You'd make your life easier if you wrote earlier on, because then the s disappear completely.

Everything up to is correct. From here, use my above advice, and the correct answer is about 2 lines away.

In fact everything is right up until your "simplifying further" bit. You can't substitute , since the 'x' here refers to the tangent line rather than the curve.
you mind helping me with another question?

The point P(at^2 , 2at) lies on the parabola C with equation y^2 = 4ax, where a is a positive constant.

(a) Show that an equation of the tangent to C at P is ty = x + at^2 (done)

The tangent to C at the point A and the tangent to C at the point B meet at the point with coordinated (-4a,3a).

(b) Find in terms of a, the coordinates of A and the coordinates of B
any ideas where to start on this one?

Turn on thread page Beta
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: September 30, 2010
Today on TSR

Been caught plagiarising...

...for the 2nd time this year

Mum says she'll curse me if I don't go to uni

Discussions on TSR

• Latest
Poll
Useful resources

Make your revision easier

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

How to use LaTex

Writing equations the easy way

Study habits of A* students

Top tips from students who have already aced their exams

Create your own Study Planner

Never miss a deadline again

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE