Verified answer

a) when x =2 y=c let R be the point (2,c)

gradient of tangent at any point is dy/dx= - c(x-1)^2 when x=2 dy/dx = -c

so gradient of normal at R =1/c equation of normal atR is y-c=(1/c)(x-2) ie yc - c^2 =x - 2

so y = x/c +c -2/c is the normal.

b) this normal meets x- axis at P. so y coordinate of P is 0, so when y=0 x=(2-c^2) so P = ( (2-c^2), 0 )

this normal meets y - axis at Q. so x coordinate of Q is 0, so when x=0 y =(c^2-2)/c so

Q = (0, (c^2 -2)/c )

c) when y-coordinate of Q is 1, (c^2 -2)/c=2 this gives c^2-2c-2=0 c= 1+root3 or 1-root3

substitute these values to get 2 different coordinates of P

I don't have time to work this out for you, but here's how you do it:

take the derivative of y = c/(x - 1) (function of a function rule)

insert the value x = 2

this gives you the slope m of the tangent to the curve at the point x = 2, y = c

the slope of the normal is therefore -1/m and its equation will be y = (-1/m)x + b

get the value of b by inserting the coordinates (2, c)

find the coordinates of P and Q by setting y = 0 and x = 0 respectively

now do part c)