If the line \(y=4 x-5\) touches the curve \(y^{2}=a x^{3}+b\) at the point \((2,3)\), then

Line \(y=4 x-5\) has slope 4
We have \(y^{2}=a x^{3}+b \Rightarrow 2 y \frac{d y}{d x}=3 a x^{2}\)
\(
\begin{array}{l}
\therefore \frac{\mathrm{dy}}{\mathrm{d} \mathrm{x}}=\frac{3 \mathrm{ax}^{2}}{2 \mathrm{y}} \\
\Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(2,3)}=\frac{3 \mathrm{a}(2)^{2}}{2(3)}=2 \mathrm{a}
\end{array}
\)
As per given condition, \(2 \mathrm{a}=4 \Rightarrow \mathrm{a}=2\)
Now \(y^{2}=a x^{3}+b\) at point \((2,3)\) becomes
\(
9=8(2)+b \Rightarrow b=-7
\)
This problem can be alternatively solved as follows:
Put \(y=4 x-5\) in \(y^{2}=a x^{3}+b\)
\(\therefore(4 x-5)^{2}=a x^{3}+b\)
When \(x=2\), we get
\(9=8 a+b$$\ldots(1)\)
Now we will go by options
Put \(\mathrm{a}=2, \mathrm{~b}=-7\) in equation \((1)\) we get
\(8(a)-7=16-7=9\)