A tangent drawn to the curve y = f (x) at P (x, y) cuts the x-axis and y-axis at A and B respectively such that BP : AP = 3 : 1 given that f (1) = 1, then

1) Equation of curve is x (dy / dx) – 3y = 0

2) Normal at (1, 1) is x + 3y = 4

3) Curve passes through (2, 1 / 8)

4) Equation of curve is x (dy / dx) + 3y = 0

5) 3 and 4

Solution: (5) 3 and 4

Equation of the tangent is Y – y = (dy / dx) (x – x)

BP / AP = 3 / 1 so that dx / x = – dy / 3y

x (dy / dx) + 3y = 0

ln x = – 1 / 3 ln y – ln c

ln x3 = – (ln cy)

(1 / x3) = cy

Given f (1) = 1

c = 1

y = 1 / x3

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