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Mathematics · 2022

JEE Main · 24 June 2022, Shift 2 · Q19

The slope of normal at any point (x, y), x>0, y>0 on the curve y=y(x) is given by x^2/(x y-x^2 y^2-1). If the curve passes through the point (1,1),…

The slope of normal at any point $\displaystyle (x, y), x>0, y>0$ on the curve $\displaystyle y=y(x)$ is given by $\displaystyle \frac{x^2}{x y-x^2 y^2-1}$. If the curve passes through the point $\displaystyle (1,1)$, then $\displaystyle e \cdot y(e)$ is equal to
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.