JEE Main 2022 (Online) 27th July Morning ShiftQuestion 1
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 ________.
Differentiation and its Application
Answer$2$
Solution
Concept. Horizontal tangent: numerator of $y'$ is zero; vertical tangent: denominator is zero. Each time, the point must also be on the curve. Formulas used. (1) $5y^4y'-9y-9xy'+2=0$ $\Rightarrow$ $y'=\dfrac{9y-2}{5y^4-9x}$Step 1 - Horizontal. By (1), $9y=2$. The curve reads $y^5-x(9y-2)=0$, so $y^5=0$, which contradicts $y=\frac29$. So $M=0$. Step 2 - Vertical. $9x=5y^4$. On the curve: $y^5-5y^5+\frac{10}{9}y^4=0$, i.e. $y^4\left(\frac{10}{9}-4y\right)=0$. So $y=0$ (point $(0,0)$) or $y=\frac{5}{18}$ (one point). At both, $9y-2\ne0$, so the tangent is vertical. $N=2$. Step 3 - Value. $M+N=2$. Answer: 2