Question

y^=ax^3at ( a,a) find the equation of tanjent and normal to the curve​

Answer 1

Answer:

Given curve is y4=ax3

On differentiating with respect to x, we get

4y3dxdy=3ax2

⇒(dxdy)(a,a)=4a33a3=43

∴ Equation of normal at point (a,a) is

y−a=−34(x−a)

⇒4x+3y=7a