JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Application of Derivatives

The line $x+90y+2=0$ has slope $-\frac{1}{90}$; for the normal at a point to be parallel to it, the normal has slope $-\frac{1}{90}$, so the tangent there has slope $90$. Differentiating, and setting this equal to $90$ gives $270x^4-540x^3-210x^2+360x+120=0$, i.e. $9x^4-18x^3-7x^2+12x+4=0$. This quartic factors with the four real distinct roots $x=1,2,-\frac{1}{3},-\frac{2}{3}$, so the curve has four such points.