If the normal to the curve ${\mathrm{y}}^2=5\mathrm{x}-1$ at the point $\left(1,-2\right)$ is of the form $\mathrm{ax}-5\mathrm{y}+\mathrm{b}=0$, then a and b are

At $\left(1, -2\right), \frac{\mathrm{dy}}{\mathrm{dx}}={\left(\frac{5}{2\mathrm{y}}\right)}_{\left(1,-2\right)}=\frac{-5}{4}$

$\therefore$equation of normal at the point $\left(1,-2\right)$ is

$$\begin{gathered} \therefore 4\mathrm{x}-5\mathrm{y}-14=0 \mathrm{given} \mathrm{ax}-5\mathrm{y}+\mathrm{b}=0 ---\left(2\right) \\ \mathrm{on} \mathrm{comparing} \left(1\right) \mathrm{and} \left(2\right) \mathrm{we} \mathrm{get} \mathrm{a} \& \mathrm{b} \end{gathered}$$