The equation of the curve passing through the point (1, 1) such that the slope of the tangent at any point (x, y) is equal to the product of its co-ordinates is
A
2logy=x2−1
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B
2logx=y2−1
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C
2logx=y2+1
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D
2logy=x2+1
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Solution
The correct option is A2logy=x2−1 According to the question dydx=xy⇒1ydy=xdx On Integrating both sides, We get- logy=x22+c Curve passes through (1,1) log1=12+c 0=12+c ⇒c=−12 logy=x22−12 2logy=x2−1