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Question
Find the equation of a curve passing through the point (1, 1). If the tangent drawn at any point P(x, y) on the curve meets the co-ordinate axes at A and B such that P is the mid-point of AB.
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Solution
Let P (x, y) be any point on the curve and AB be the tangent to the given curve at P.
P is the midpoint of AB .....(Given)
∴ Coordinates of A and B are (2x, 0) and (0, 2y) respectively.
∴ Slope of the tangent AB = `(2y - 0)/(0 - 2x) = - y/x`
∴ `("d"y)/("d"x) = - y/x`
⇒ `("d"y)/y = -("d"x)/x`
Integrating both sides, we get
`int ("d"y)/y = -int ("d"x)/x`
⇒ log y = – log x + log c
⇒ log y + log x = log c
⇒ log yx = log c
∴ yx = c
Since, the curve passes through (1, 1)
∴ 1 × 1 = c
∴ c = 1
Hence, the required equation is xy = 1.
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