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प्रश्न
The Slope of the tangent to the curve at any point is the reciprocal of four times the ordinate at that point. The curve passes through (2, 5). Find the equation of the curve
उत्तर
The slope of the tangent to the curve at any point is
= `1/(4("ordinate at the point"))`
`("d"y)/("d"x) = 1/(4y)`
The equation can be written as
4y dy = dx ........(1)
Integrating equation (1) on both sides, we get
`4int y "d"y = int "d"x`
`4("y"^2/2)` = x + c
2y2 = x + c …….. (2)
Since the curve passes through at (2, 5), we get
2(5)2 = 2 + c
50 = 2 + c
50 – 2 = c
48 = c
∴ Substituting the value of c in equation (2), we get
2y2 = x + 48 is the required equation of the curve.
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