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Question
Find the curve whose gradient at any point P(x, y) on it is `(x - "a")/(y - "b")` and which passes through the origin
Solution
The gradient of the curve at P(x, y)
`("d"y)/("d"x) = (x - "a")/(y - "b")`
(y – b) dy = (x – a) dx
Integrating on both sides
`int (y - "b") "d"y = int (x - "a") "d"x`
⇒ `(y -- "b")^2/2 = (x - )^2/2 + "c"`
Multiply each term by 2
∴ (y – b)2 = (x – a)2 + 2c .........(1)
Since the curve passes through the origin (0, 0)
Equation (1)
(0 – b)2 = (0 – a)2 + 2c
b2 = a2 + 2c
b2 – a2 = 2c .......(2)
Substitute equation (2) and (1)
(y – b)2 = (x – a)2 + b2 – a2
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