The differential equation for all the straigh | vedclass

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(C) The equation of a straight line at a unit distance from the origin is given by $x \cos \alpha + y \sin \alpha = 1$ $... (i)$Differentiating with r

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${\left( {y - x\frac{{dy}}{{dx}}} \right)^2} = 1 + {\left( {\frac{{dy}}{{dx}}} \right)^2}$

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(C) The equation of a straight line at a unit distance from the origin is given by $x \cos \alpha + y \sin \alpha = 1$ $... (i)$Differentiating with respect to $x$,we get $\cos \alpha + \frac{dy}{dx} \sin \alpha = 0$ $... (ii)$From $(ii)$,we have $\cos \alpha = -\frac{dy}{dx} \sin \alpha$. Substituting this into $(i)$:$x(-\frac{dy}{dx} \sin \alpha) + y \sin \alpha = 1$$\sin \alpha (y - x \frac{dy}{dx}) = 1$$y - x \frac{dy}{dx} = \frac{1}{\sin \alpha} = \csc \alpha$ $... (iii)$From $(ii)$,$\frac{dy}{dx} = -\cot \alpha$,so $(\frac{dy}{dx})^2 = \cot^2 \alpha$.Using the identity $1 + \cot^2 \alpha = \csc^2 \alpha$,we substitute from $(iii)$ and $(iv)$:$1 + (\frac{dy}{dx})^2 = \csc^2 \alpha = (y - x \frac{dy}{dx})^2$.

The differential equation for all the straight lines which are at a unit distance from the origin is

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