The slope of normal at any point ( {\x}, {\y}), {\x}>0, {\y}>0 on the curve {\y}= {\y}( {\x}) is given by \frac{x^2}{x y-x^2 y^2-1}. If the curve passes through the point (1,1), then e. y(e) is equal to [24-Jun-2022-Shift-2]

Correct Answer is : 1+tan(1)1−tan(1)

JEE Main 24 June 2022 Shift 2 Solved Paper

Section: Mathematics

Question : 19 of 90

Marks: +1, -0

The slope of normal at any point (x,y),x>0,y>0 on the curve y=y(x) is given by ‌

xy−x2y2−1

. If the curve passes through the point (1,1), then e. y(e) is equal to

[24-Jun-2022-Shift-2]

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