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 $8(a)-7=16-7=9$