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If a curve is such that line joining origin to any point P(x, y) on the curve and line parallel to y–axis through P are equally inclined to tangent to curve at P, then the differential equation of the curve is

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

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

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

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

answer is A.

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Detailed Solution

$θ_{1} + α = θ_{2}, θ_{1} + 2 α = \frac{π}{2}, θ_{1} + 2 (θ_{2} - θ_{1}) = \frac{π}{2}$ , $2 θ_{2} - θ_{1} = \frac{π}{2}$

$\frac{2 \tan (θ_{2})}{1 - {(\tan (θ_{2}))}^{2}} = \frac{1}{\tan θ_{1}}, \frac{2 \frac{d y}{d x}}{1 - {(\frac{d y}{d x})}^{2}} = - \frac{x}{y}$

$2 y \frac{d y}{d x} = + x {(\frac{d y}{d x})}^{2} - x \Rightarrow x {(\frac{d y}{d x})}^{2} - 2 y \frac{d y}{d x} = x$

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