Question

since the equation of lines whose distance from origin is unit, is given by
[
x cos alpha+y sin alpha=1
]
Differentiate w.r.t ( x, ) we get ( cos alpha+frac{d y}{d x} sin alpha=0 )
On eliminating the ' ( { }^{prime} alpha^{prime} ) with the help of (i) and (ii)
i.e.,
(i) ( -x times(text { ii }) )
( Rightarrow quad sin alphaleft(mathrm{y}-mathrm{x} frac{mathrm{dy}}{mathrm{dx}}right)=1 Rightarrowleft(mathrm{y}-mathrm{x} frac{mathrm{dy}}{mathrm{dx}}right)=operatorname{cosec} alpha )
(iii)
Also
( (i i) Rightarrow frac{d y}{d x}=-cot alpha Rightarrowleft(frac{d y}{d x}right)^{2}=cot ^{2} alpha )
Therefore by (iii) and (iv), ( 1+left(frac{d y}{d x}right)^{2}=left(y-x frac{d y}{d x}right)^{2} )

# The differential equation for all the straight lines which are at a unit distance from the origin is (A)(y-kay--( V- (y+medy y +X Cebu 13011511626|

Solution