Parametric question - help?

The parametric equations of a curve are x = 2t^2 and y = 4t.

Two points on the curve are P(2p^2, 4p) and Q(2q^2, 4q)

i) Show that the gradient of the normal to the curve at P is −p

I solved the differential to be 1/t, and so I would have thought that the negative reciprocal is -t, but why is it -p? if anything is should be -2p^2 ??

The parametric equations of a curve are x = 2t^2 and y = 4t.

Two points on the curve are P(2p^2, 4p) and Q(2q^2, 4q)

i) Show that the gradient of the normal to the curve at P is −p

I solved the differential to be 1/t, and so I would have thought that the negative reciprocal is -t, but why is it -p? if anything is should be -2p^2 ??

$\displaystyle \frac{\mathrm{d}y}{{d}x} = \frac{4}{4t} = \frac{1}{t} = \frac{4}{y}$.

So the gradient of the normal to the curve at P is the negative reciprocal of the gradient evaluated at the point P.

So $\frac{4}{4p} = \frac{1}{p}$ so normal = -p.

since y=4p at P.

Bro, you can't just give him the working..

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