Question

find the equation of tangent and equation of normal to the curve y⁴=ax³ at (a,a)

answer fastly☺...

Answer 1

y⁴=ax³

by differentiating with respect to x

4y³dy/dx=3ax²

dy/dx=3ax²/4y³


dy/dx at (a,a)is 3/4


equation of tangent =

(y-y')=m(x-x')

y-a=3/4(x-a)

4y-4a=3x-3a

3x-4y+a=0 is the required tangent equation


slope of normal will be -dx/dy=-4/3

so equation of normal is (y-y')=m(x-x')

(y-a)=-4/3(x-a)

3y-3a=4a-4x

3y+4x+a=0 is the equation of normal



I hope this will help u ;)