a curve ahs parametric equations x=t^2

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.

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.

To find the gradient of the curve at a specific point, you need to first differentiate both the x and y equations with respect to t. Let's differentiate the given equations:

Given:
x = t^2
y = 1 - (1/2)t

Differentiating x with respect to t:
dx/dt = 2t

Differentiating y with respect to t:
dy/dt = -1/2

To find the gradient at the point P(1/4, 0), substitute t = 1/4 into the gradients:

Gradient of x: dx/dt = 2(1/4) = 1/2
Gradient of y: dy/dt = -1/2

Therefore, the gradient of the curve at point P is 1/2.