
6-46 Calculus – Differentiation and Integration
Example 6.36 Find the coordinates of the centre of curvature of the parabola y
2
= 4ax at any point
(x, y) on the curve.
Solution: Given equation of the parabola is y
2
= 4ax.
Differentiating w.r.t x, we get
2 4y
dy
dx
a= .
Thus,
dy
dx
a
y
a
ax
a
x
a x= = = = ⋅
−
2 2
4
1 2/
d y
dx
a x x
y
a
d y
dx
a
y
a
2
2
3 2
2
2
2
2
1
2 4
1
2 4
= − ⋅ ⋅ =
∴ = − ⋅ ⋅
− /
But
−−
−
= ⋅ ⋅
= − ⋅ =
−
3 2
3 3 2
3
2
3
2
4
2
8 4
/
/
( )
a
y a
a a a
y
a
y
The coordinates of centre of curvature are given by
a = −
+
= −
+
x
dy
dx
dy
dx
d y
dx
x
a
y
a
y
1
2
1
4
2
2
2
2
2
−
= +
+
=
+ +
=
+
= +
4
4
2
2 4
2
6 4
2
3
2
3
2 2 2 2 2
a
y
x
y a
a
y ax a
a
ax a
a
x 22a.
b = +
+
= +
+
−
= −
+