Very difficult parametric question ??

for A i tried to work out solution for t by equaling f'(x) to 2 and I got (t^2-9) difference of two squares t= +/- 3 now I can't seem to work from here as I don't think this is the correct solution
plsss
help ??

Well, that's what I just got, too, so why not just substitute those values back into the values for x and y?

Why don't you think those values of t are correct?

Got part A how do I continue ??

for A i tried to work out solution for t by equaling f'(x) to 2 and I got (t^2-9) difference of two squares t= +/- 3 now I can't seem to work from here as I don't think this is the correct solution
plsss
help ??

this is the solution to b - i don't knwo how to solve for t

this is the solution to b - i don't knwo how to solve for t

The cartesian equation cannot be expressed in the explicit form "y=..." since there are either 0,1, or 2 possible values for y for a given x value. You need to leave it in implicit form.

b) is just to substitute for x^2; write it in terms of t (repeat for y).

my attempt at part C don't know how to do ii

I'm just going to assume you did all this correctly since I don't have time to check your working.

For part c(ii) you seek the second point of intersection between the normal, which is the line $y+2 = \dfrac{1}{2}(x-4)$, and the original curve which is defined parametrically.

If both equations were Cartesian, I am sure you would know how to proceed and find points of intersection. The whole 'sub y from one equation into the other and solve for x' type of thing.

But since one eqn is Parametric, and the other is Cartesian, you still aim to end up solving a single equation in one variable.

Your Cartesian line has x,y in there but these can both be expressed in terms of the parameter t.

Hence, if you sub in the curve's defining x,y coordinates in terms of t into the normal line equation, you end up with one equation in t to solve.

If you got the correct normal, then t=1 is obviously one solution corresponding to the point of intersection where you constructed the normal.

You seek *different* solution for t, and once you have it you can sub it back into the curve's eqn to obtain the x,y coordinates.

thank you for your reply
subbed into normal eq I got t^2-9 = 0 which means that t= -3 or t= 3 which is the values I got for t in the beginning part of the question and subbing these into eq for x and y will give me the same coordinates super confused ??