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March 28, 2015

March 28, 2015

Posted by **Hannah** on Tuesday, February 22, 2011 at 5:24am.

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.

- Maths -
**Reiny**, Tuesday, February 22, 2011 at 9:21amyou 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.

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