If area of triangle formed by tangent (with positive slope) and normal to a curve at any point in first quadrant with x-axis is cube of its ordinate, then differential equation of such family is -

{(\frac{d y}{d x})}^{2} = 2 y \frac{d y}{d x}

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{(\frac{d y}{d x})}^{2} - 2 y \frac{d y}{d x} + 1 = 0

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2 y (\frac{d y}{d x}) + 1 = 0

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{(\frac{d y}{d x})}^{2} + 2 y \frac{d y}{d x} + 1 = 0

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Solution

The correct option is B ${(\frac{d y}{d x})}^{2} - 2 y \frac{d y}{d x} + 1 = 0$
Area = $\frac{1}{2} (S T + S N) \times P S$
$\Rightarrow y^{3} = \frac{1}{2} (y y^{'} + \frac{y}{y^{'}}) . y$
$\Rightarrow 2 y = y^{'} + \frac{1}{y^{'}}$
$\Rightarrow 2 (\frac{d y}{d x})^{2} - 2 y (\frac{d y}{d x}) + 1 = 0$

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Similar questions

The differential equation of family of curves $y^{2} = 4 a (x + a)$ is
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d) $2 x \frac{d y}{d x} + y {(\frac{d y}{d x})}^{2} - y = 0$

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If area of triangle formed by tangent (with positive slope) and normal to a curve at any point in first quadrant with x-axis is cube of its ordinate, then differential equation of such family is -

The correct option is B (dydx)2−2ydydx+1=0Area = 12(ST+SN)×PS⇒y3=12(yy′+yy′).y⇒2y=y′+1y′⇒2(dydx)2−2y(dydx)+1=0