Answer: \(a=-1\), \(b=\frac14\).

Answer: \(a=-1\), \(b=\frac14\).

The curve is

\(y=\frac ax+3x-2.\)

Differentiate:

\(\frac{dy}{dx}=-\frac{a}{x^2}+3.\)

At \(x=1\), the gradient of the tangent is

\(3-a.\)

The normal has gradient \(-\frac14\), so the tangent has gradient \(4\). Hence

\(3-a=4.\)

Therefore

\(a=-1.\)

When \(x=1\), the point on the curve has

\(y=a+3-2=a+1=0.\)

So the normal passes through \((1,0)\). Its equation is

\(y=-\frac14x+b.\)

Substitute \((1,0)\):

\(0=-\frac14+b.\)

Thus

\(b=\frac14.\)