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# Co-ordinate Geometry. watch

1. Hello,

I really can't get my head around co-ordinate geometry in C4. Could someone help me with the following question:

A curve C has parametric equations x = at^2, y = 2at. Show that the equation of the normal to C at the point P, whose parameter is P, is

px + y - 2ap - ap^3 = 0.

I dont have a clue where to start. Where does P come into the equation? I started by differentiating x and y and then got dy/dx to be t.

Could someone give me a hint? Also I dont understand "whose parameter is P".

Thanks.
2. At the point P, the parameter is p so t=p. This means that at point P, x=ap^2 and y=2ap.

dy/dx= (dy/dt) / (dx/dt) = 1/t so normal gradient is -t.
Gradient of normal at the point P is therefore -p.
Then use y-y1=m(x-x1) with point P to find equation of normal.
3. (Original post by vc94)
At the point P, the parameter is p so t=p. This means that at point P, x=ap^2 and y=2ap.

dy/dx= (dy/dt) / (dx/dt) = 1/t so normal gradient is -t.
Gradient of normal at the point P is therefore -p.
Then use y-y1=m(x-x1) with point P to find equation of normal.
Oh I see. Thanks.

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