The curve is described by two functions:

Find the derivative, which will be, for the positive square root function:

y' is not defined at x = 2a because the function is not differentiable at this point (although it is continuous). Nevertheless you can find the limits of y' as

and

which will be different due to the absolute value function in the denominator.

Of course, this is the gradient of the tangent, so the gradient of the normal will be

You can find the equation for the normal line using:

Now it should be obvious how to show that the normal line intersects the positive square root function at the given point.