Example 34 Find 
2
2
2
3
3
3
dy, if
x
y
a
dx
+
= Solution Let x = a cos3 θ, y = a sin3 θ Then
2
2
3
3
x
+y
 =
2
2
3
3
3
3
( cos
)
( sin
)
a
a
θ
+
θ
=
2
2
2
2
3
3
(cos
(sin
)
a
a
θ +
θ =
Hence, x = a cos3θ, y = a sin3θ is parametric equation of 
2
2
2
3
3
3
x
y
a
+
=
Now
dx
dθ  = – 3a cos2 θ sin θ and dy
dθ  = 3a sin2 θ cos θ
Therefore
dy
dx  =
2
3
3 sin2
cos
tan
3 cos
sin
dy
a
y
d
dx
x
a
d
θ
θ
θ =
= −
θ = −
−
θ
θ
θ
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
137
EXERCISE 5 6
If  x and y are connected parametrically by the equations given in Exercises 1 to 10,
without eliminating the parameter, Find dy
dx