The question is phrased as follows:
"A rectangular hyperbola, W, has equation xy = 12"
a) Show that the gradient of the normal, N, to W at the point P(2,6), is 1/3.
b) Hence find an equation for N/
c) Find the coordinates of the point Q where N intersects the curve W again.
My solution for a) & b) was the linear equation 3y = x + 16, but using the information provided I'm quite lost as to the other point of intersection...a simplified solution would be met with much gratification!
You found the equation of the normal line $\:y=\frac{x+16}{3}\:$, which is correct.
In order to find the points of intersection with the hyperbola $\:y=\frac{12}{x}\:$ simply write that both $\:y\:$ are equal : $$y=\frac{x+16}{3}=\frac{12}{x}$$ $$x^2+16x-36=0$$ Solve it for $\:x\:$. You will found two roots :
First $\:x=2\:$ is obvious since it is already known that the normal line is issued from the point $\;(x=2\:,\:y=6)\:$
Second $\:x=-18\:$ and then $\:y=\frac{12}{x}=-\frac{2}{3}$