The equation of curve in which portion of y-axis cutoff between origin and tangent varies as cube of abscissa of point of contact is

Options

A. ` y = (kx^(3))/3 + Cx`
B. ` y - (kx^(2))/2 +C`
C. ` y = -(kx^(3))/2 + Cx`
D. ` y = (kx^(3))/3 + (Cx^(2))/2`

Answer

c

Explanation

Equation of tangent at (x,y) is
` Y - y = (dy)/(dx) (X-x)`
For Y - axis ,X = 0
Then , ` Y = y - x (dy)/(dx)`
Given , ` (y - x (dy)/(dx)) prop x^(3)`
` rArr y - x (dy)/(dx) = kx^(3) rArr (dy)/(dx) - y = - kx^(2)`
which is linear differential equation
` :. IF = e^(-int1//xdx) = e^(-"inx")= e^("In"(1//x))= 1/x `
Then, solution is ` y(1/x)=int (-kx^(2))/x dx+C`
` rArr y/x = -(kx^(2))/2 +C or y = -(kx^(3))/2 + Cx`