If the normal to the curve $y^2=5x –1$ at the point $(1, –2)$ is of the form $ax–5y+b=0$, then $a$ and $b$ are

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

$\therefore$Equation of normal at the point (1,–2) is

$\therefore 4x-5y-14=1\text{. given }ax-5y+b=0-\left(2\right)\text{ on }$

comparing (1) & (2) we get a & b