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The slope of the normal at any point (x, y), with x > 0 and y > 0, on a curve y = y(x) is x²/(xy - x² y² - 1). If the curve passes through (1, 1), find the value of e * y(e).
- (1 - tan(1))/(1 + tan(1))
- tan(1)
- 1
- (1 + tan(1))/(1 - tan(1))
Correct answer: (1 + tan(1))/(1 - tan(1))
Solution
The substitution v = xy gives arctan(xy) = ln x + pi/4, so e*y(e) = tan(1 + pi/4) = (1 + tan 1)/(1 - tan 1).
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