Find the equation of the normal at the point `(a m^2,a m^3)`for the curve `a y^2=x^3`.

Options

A. `2x+3my-3am^(3)-2am^(2)=0`
B. `2x+3my-3am^(4)-2am^(2)=0`
C. `2x+3m^(2)y-3am^(3)-2am^(3)=0`
D. None of the above

Answer

B

Explanation

Given `ay^(2)=x^(3)`……………..i
`impliesa(2y)(dy)/(dx)=3x^(2)implies(dy)/(dx)=(3x^(2))/(2ay)`
The slope of the tangent to the given curve at `(am^(2),am^(3))` is
`((dy)/(dx))_((am^(2),am^(3)))=(3(am^(2))^(2))/(2a(am^(3)))=3/2m`
`:.` Slope of normal at `(am^(2),am^(3))=-1/(3/2m)=-2/(3m)`
Hence, equation of normal at `(am^(2),am^(3))` is given by
`y-am^(3)=-2/(3m)(x-am^(2))`
`implies3my-3am^(4)=-2x+2am^(2)`
`implies2x+3my-3am^(4)-2am^(2)=0`