Curves C1 and C2 have equations y= ln(4x-7)+18 and y= a(x^2 +b)^1/2 respectively, where a and b are positive constants. The point P lies on both curves and has x-coordinate 2. It is given that the gradient of C1 at P is equal to the gradient of C2 at P.

y= ln(4x-7)+18 y= a(x^2 +b)^1/2
At x=2 dy/dx = dy/dx and y =y
At x =2 y = ln(8-7) +18 y = ln 1 +18 y =18
At x = 2 18=a(4 +b)^1/2 18/(4+b)^1/2= a
y=ln(4x-7)+18dy/dx= 4/(4x-7)
At x= 2 dy/dx= 4/8-7= 4
y=a(x^2 +b)^1/2 dy/dx= xa(x^2 +b)^-1/2
At x=2 4=2a(4+b)^-1/22=a(4+b)^ -1/22(4+b)^1/2= a
2(4+b)^1/2= a 18/(4+b)^1/2= a 18/(4+b)^1/2= 2(4+b)^1/29=4+b5=b
2(4+b)^1/2= a 2(4+5)^1/2= a 2(9)^1/2= a 2(3)=a6=a
a=6 b=5