Normal on the point of curve

Question

I tried to determine slope of curve at any variable point write the equation of normal but i am stucked what should i do?

Answer is 4

Answer 1

$3y=6x-5x^3$

$\dfrac{dy}{dx}=2-5x^2$

If two lines are perpendicular and they have slopes $m_1,m_2$ then $m_1=\frac{-1}{m_2}$

Slope of normal $m= \dfrac{-1}{2-5x^2}$

Question says this line passes through origin . Equation will be y=mx

$y=\dfrac{-x}{2-5x^2}$

$x+2y=5x^2y$

Now put option values of x,y in this equation,if LHS=RHS then it is your correct option.