MEI Maths Item of the Month
October 2015
AOB
The points A and B are on the curve y = x² such that AOB is a right angle. What points A and
B will give the smallest possible area for the triangle AOB?
Solution
Writing the coordinates of the points A and B as (a, a 2 ) and (b, b 2 ) gives the line OA
gradient a.
1
.
a
1
1
 1 1 
Solving b 2   b gives b   and hence   , 2  for the coordinates of B.
a
a
 a a 
The line OB has gradient: 
By dropping perpendiculars to the x-axis, and labelling these points C and D, the area of the
triangle AOB can be found by subtracting the area of two triangles from a trapezium:
1 of 2
TB v1.0 © MEI
30/03/2016
MEI Maths Item of the Month
AOB = CABD  OAC OBD
1
 2 1 
1
 a  2  a   a 3
a
a

   a3

2
2
2
1 1
1
a3  a   3  a3  3
a a
a

2
1
a
a

2
Solving with calculus:
A
1
1
1 2
d
A
a,
a .

2
da
2
a
dA
 0 when a  1 (or a  1 ). This gives A and B at (1,1) and (1,1) .
da
Solving without calculus:
2
1  a  1  2
a


a
a 
which has a minimum value when
2
2
(or a  1 ). This gives A and B at (1,1) and (1,1) .
2 of 2
a
1
 0 , i.e. when a  1
a
TB v1.0 © MEI
30/03/2016