If the length of the sub tangent at any point p (x ,y ) on a curve f (x ,y )=0 is x+7y^2, then f (x ,y )=

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Accepted Answer

Correct answer: C. (c) Given, f (x ,y )=0 We know that, Length of subtangent = y dy dx ⇒x+7y^2= y dy dx ⇒ dy dx= yx+7y^2 ⇒ dx dy= x+7y^2y ⇒ dx dy- 1yx=7y ∴ IF=e^ P (y ) dy=e^ - 1y dy=e^- log ⁡y= 1y Here, P (y )= -1y Q (y )=7y So, solution is given by x 1y= 7y 1y dy+C ⇒ xy=7y+Cx=7y^2+ cy ⇒7y^2+ cy-x=0 ∴f (x ,y ) :7y^2+ cy-x=0

$If the length of the sub tangent at any point p (x, y) on a curve f (x, y) = 0 is x + 7 y^{2}, then f (x, y) =$

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