JEE Math Practice Question
The slope of normal at any point (x, y), x > 0, y > 0 on the curve y = y(x) is given by $${{{x^2}} \over {xy - {x^2}{y^2} - 1}}$$. If the curve passes through the point (1, 1), then e . y(e) is equal to
- A.$${{1 - \tan (1)} \over {1 + \tan (1)}}$$
- B.tan(1)
- C.1
- D.$${{1 + \tan (1)} \over {1 - \tan (1)}}$$
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