Answer: \(P=\left(0,\ln6-\frac{2}{2\ln6-\frac23}\right)\), approximately \((0,1.11)\).
The key point is to differentiate first, then substitute the required value or solve the resulting equation.
Let
\(y=\frac{\ln(x^2+2)}{2x-3}.\)
Using the quotient rule,
\(\frac{dy}{dx} =\frac{(2x-3)\frac{2x}{x^2+2}-2\ln(x^2+2)}{(2x-3)^2}.\)
At \(x=2\),
\(y=\frac{\ln6}{1}=\ln6.\)
The gradient of the tangent is
\(\frac{dy}{dx} =\frac{1\cdot\frac46-2\ln6}{1^2} =\frac23-2\ln6.\)
Therefore the gradient of the normal is
\(-\frac1{\frac23-2\ln6} =\frac1{2\ln6-\frac23}.\)
The normal passes through \((2,\ln6)\), so its equation is
\(y-\ln6=\frac1{2\ln6-\frac23}(x-2).\)
At the \(y\)-axis, \(x=0\). Hence
\(y-\ln6=-\frac2{2\ln6-\frac23}.\)
So
\(y=\ln6-\frac2{2\ln6-\frac23}.\)
Thus
\(P=\left(0,\ln6-\frac2{2\ln6-\frac23}\right).\)
Numerically,
\(P\approx(0,1.11).\)
