Maths problem!!! Pls Help!

(i) x=t-1, y=t^2-2t
y=(t^2-2t+1)-1
y=[(t-1)^2]-1
y=x^2-1----------(1)
x^2=y+1

(ii) x^2=y+1
y=x^2-1
dy/dx=d(x^2-1)/dx
dy/dx=2x----------(2)
(dy/dt)/(dx/dt)=2(t-1)------(3)
gradient of the tangent to this parabola at the point with parameter t
=2(t-1)

(iii) it is easy and direct to use the equations (1) and (2) with variables x,y
if you use equation (3),
you need to responding value of t for finding y and dy/dx.
more steps to be needed for t but you can it for giving the same answer.
P(0) means x=0,
substitute x=0 into (1), y=[(0)^2] -1= -1
substitute x=0 into (2), dy/dx=2(0)=0
the equation of the tangent to the parabola at the point P(0) is
[y- (-1)]/[(x- (0)]=0
(y+1)/x=0
y+1=0
or y= -1

P(2) means x=2,
substitute x=2 into (1), y=[(2)^2] -1=3
substitute x=2 into (2), dy/dx=2(2)=4
the equation of the tangent to the parabola at the point P(2) is
[y- (3)]/[(x- (2)]=4
y -3=4(x -2)
y -3=4x -8
y=4x -5
or 4x -y -5=0