Find the equation of the curve in which the perpendicular from the origin on any tangent is equal to the abscissa of the point of contact.

Options

A. The required differential equation is `y^(2) - 2xy. (dy)/(dx) - x^(2) = 0`
B. It is not homogeneous differential equation
C. The solution of the differential equation is `x^(2)+y^(2)=C x`
D. The solution of the differential equation is `2x^(2) +3y^(3) = C xy`

Answer

c

Explanation

The tangent at ny point P(x,y) of the curve is
` Y - y - (dy)/(dx) (X-x)=0`
Length of perpendicular from origin ` (x(dy)/(dx)-y)/(sqrt([1+((dy)/(dx))^(2)]))=x`
`" [given tangent at point P of the curve]"`
On squaring both sides, we get
`x^(2)((dy)/(dx))^(2)-2xy(dy)/(dx)+y^(2)=x^(2)((dy)/(dx))^(2)`
`rArr" "y^(2)-2xy(dy)/(dx)-x^(2)=0`
`rArr" "(dy)/(dx)=(y^(2)-x^(2))/(2xy)" [homogeneous differential equation]"`
` rArr 2xy (dy)/(dx)- y^(2) = -x^(2) `
` rArr 2y (dy)/(dx) -(y^(2))/x = -x`
` rArr 2y (dy)/(dx) -y^(2)= -x^(2)`
Put ` y^(2) = v rArr 2y(dy)/(dx) = (dv)/(dx)`
` rArr (dv)/(dx) - v/x = -x ` [ linear differetnial equation ]
` :. IF = e^(int - 1/x dx) = e^(log.(1)/(x)) = 1/x `
Solution of linear differetial equation ,
` v*1/x = int -x *1/x dx+C= - x +C rArr (y^(2))/x = -x +C`
` rArr x^(2) + y^(2) = Cx`