Answer: The midpoint of \(AB\) is \(\left(3,\frac{15}{2}\right)\).

When \(x=4\),

\(y=\sqrt{4(4)+9}=\sqrt{25}=5.\)

So the point of contact is \((4,5)\).

Differentiate:

\(y=(4x+9)^{1/2}\) \(\frac{dy}{dx}=\frac12(4x+9)^{-1/2}\cdot4=\frac{2}{\sqrt{4x+9}}.\)

At \(x=4\),

\(\frac{dy}{dx}=\frac{2}{5}.\)

This is the gradient of the tangent, so the gradient of the normal is

\(-\frac{5}{2}.\)

The equation of the normal is

\(y-5=-\frac52(x-4).\)

To find the \(x\)-intercept, put \(y=0\):

\(-5=-\frac52(x-4).\)

So

\(-10=-5(x-4)\Rightarrow 2=x-4\Rightarrow x=6.\)

Hence \(A=(6,0)\).

To find the \(y\)-intercept, put \(x=0\):

\(y-5=-\frac52(-4)=10,\)

so \(y=15\). Hence \(B=(0,15)\).

The midpoint of \(AB\) is therefore

\(\left(\frac{6+0}{2},\frac{0+15}{2}\right)=\left(3,\frac{15}{2}\right).\)