the curve, C has the equation:

$\displaystyle 4(x+1)^2 - (y-1)^2 =1$

Find the greatest value of k, where k is a positive integer, for which the curve $\displaystyle y=ln(x+k)$ cuts C at only one point.

what i tried to do is:

$\displaystyle 4(x+1)^2 - (y-1)^2 =1$

$\displaystyle 4x^2 + 8x - y^2 +2y +2=0$

By substitution,

$\displaystyle 4x^2+8x-[ln(x+k)]^2+2[ln(x+k)]+2=0

$

and then i'm stuck because of $\displaystyle [ln(x+k)]^2$.

is the approach wrong or did i make a mistake somewhere?