Question

Let $M$ and $N$ be the number of points on the curve $y^5-9xy+2x=0$, where the tangents to the curve are parallel to $x$-axis and $y$-axis, respectively. Then the value of $M+N$ equals _____

Answer 1

Answer: 2

$y^5-9xy+2x=0$
$5y^4\frac{dy}{x}-9x\frac{dy}{dx}-9y+2=0$
$\frac{dy}{dx}\left(5y^4-9x\right)=9y-2$
$\frac{dy}{dx}=\frac{9y-2}{5y^4-9x}=0$ (for horizontal tangent) $y=\frac{2}{9}\Rightarrow$ Which does not satisfy the original equation $\Rightarrow M=0$
Now $5y^4-9x=0$ (for vertical tangent)
$5y^4(9y-2)-9y^5=0$
$y^4[45y-10-9y]=0$
$y=0(Or)36y=10$
$y=\frac{5}{18}$
$y=0\Rightarrow x=0\&y=\frac{5}{18}\Rightarrow x=$
$(0,0)\left(x,\frac{5}{18}\right)$
$N=2$
$M+N=0+2=2$