a curve ahs parametric equations x=t^2

Question

and y= 1-1/2t for t>0.

i)find the co-ordinates of the point P where the curve cuts the x-axis which i found to be P(1/4, 0)

the next part i cant do
ii) find the gradient of the curve at this point.

So far, I have the gradient to be:
-2/4t^3

is that gradient right and how do I get the value at the point P.

Answer 1

you could change it to cartesian form ...

t^2 = x
t = √x

y = 1 - (1/2)t
2y = 2 - t
t = 2 - 2y

so √x = 2 - 2y
√x + 2y = 2
to cut the x-axis, y = 0
√x=2
x=4

so the x-intercept is (4,0)

>b>Unless ... you meant to type y = 1 - 1/(2t)
in that case the point is correct and the cartesian equation would be
√x = 1/(2-2y)

(but you didn't type it that way)

In either case, differentiate the correct equation implicitly, and sub in the corresponding x-intercept.