MathematicsMedium175×since 2002Q2787The slope of normal at any point (x, y), x > 0, y > 0 on the curve y = y(x) is given by x2xy−x2y2−1{{{x^2}} \over {xy - {x^2}{y^2} - 1}}xy−x2y2−1x2. If the curve passes through the point (1, 1), then e . y(e) is equal toA1−tan(1)1+tan(1){{1 - \tan (1)} \over {1 + \tan (1)}}1+tan(1)1−tan(1)Btan(1)C1D1+tan(1)1−tan(1){{1 + \tan (1)} \over {1 - \tan (1)}}1−tan(1)1+tan(1)Check answerSkip