Question

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

Answer 1

$\frac{dy}{dx}=\left(\frac{5}{2y}\right)$
$(\frac{dy}{dx}{)}_{(1,-2)}=\frac{-5}{4}$
$\therefore$ Equation of normal at the point $(1,-2)$ is
$\therefore 4x-5y-14=0\cdots \left(1\right)$
Given, $ax-5y+b=0\cdots \left(2\right)$
On comparing $\left(1\right)$ and $\left(2\right),$ we get
$a=4$ and $b=-14$
$\therefore a-b=18$