Question

find the equation of normal to the curve x = a cos ^3 theta ,y= bsin^3 theta at theta​

Answer 1

Answer:

We have,

x=acos

3

θ,y=asin

3

θ

dθ

dx

=−3acos

2

θsinθ

dθ

dy

=3asin

2

θcosθ

Now,

dx

dy

=

−3acos

2

θsinθ

3asin

2

θcosθ

=−tanθ

Therefore, slope of the normal at any point on the curve = −

dx

dy

1

=−

−tanθ

1

=cotθ

Hence, Slope of the normal at θ=

4

π

=cot

4

π

=1