Answer: (a) \((5,4)\); (b) \(y=-\frac{10}{3}x+\frac{95}{3}\).

Answer: (a) \((5,4)\); (b) \(y=-\frac{10}{3}x+\frac{95}{3}\).

(a) Since

\(y=(1+3x)^{1/2},\)

we have

The gradient of the normal is \(-\frac83\), so the gradient of the tangent is the negative reciprocal:

\(m_{\text{tangent}}=\frac38.\)

Set the derivative equal to \(\frac38\):

\(\frac3{2\sqrt{1+3x}}=\frac38.\)

Cancel \(3\) and solve:

\(\frac1{2\sqrt{1+3x}}=\frac18, \qquad 2\sqrt{1+3x}=8.\)

Thus

\(\sqrt{1+3x}=4, \qquad 1+3x=16, \qquad x=5.\)

Then

\(y=\sqrt{1+3(5)}=4.\)

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

(b) At \((8,5)\),

The normal gradient is therefore

\(-\frac{10}{3}.\)

Using \(y-y_1=m(x-x_1)\),

\(y-5=-\frac{10}{3}(x-8).\)

Hence

\(y=-\frac{10}{3}x+\frac{95}{3}.\)