Question

find the equation of the tangent and normal to the curve y³=2ax² at the point (2am³,2am²)

Answer 1

y cube = 2ax square
Differentiate it w. r. t x
3y square dy/dx = 4ax
dy/dx = 4ax/3y square
[dy/dx]at point (2am cube,2am square) = 8am cube/12a square m power four
= 2/3am
Slope of tangent = 2/3am
Now , equation of tangent is:
(y-2a m square) = 2/3am ×(x-2am cube)
now solve this equation

Now slope of normal = -3am/2
equation of normal is:
(y-2am square) -3am/2 × (x-2am cube)
now solve this equation