Coordinate Geometry

Question

Given that the points O(0,0)and B(1/? ,1/?^2 )where ? >0,show that the area of ?AOB is (1+?)/(2?^2 ) sq units.Find AB in termsof ? ?.Show that the height from O joining the line A and B is 1/(√(2?^2-2?-1)) units and hence find the value of ? for which this height is the largest.

Answer 1

Using the formula for the area of a triangle with coordinates,

Area = (1/2)|x1y2 − y1x2|

Points:

O(0,0)

A(-1,1)

B(1/t , 1/t^2)

Area = (1/2)|(-1)(1/t^2) − (1)(1/t)|

= (1/2)|-1/t^2 − 1/t|

Since t > 0,

Area = (1/2)(1/t^2 + 1/t)

= (1/2)((1+t)/t^2)

Therefore,

Area of triangle AOB = (1+t)/(2t^2)

Finding AB:

Using the distance formula,

AB = √[(1/t + 1)^2 + (1/t^2 − 1)^2]

= √[((t+1)/t)^2 + ((1−t^2)/t^2)^2]

= √[((t+1)^2/t^2) + ((1−t^2)^2/t^4)]

Taking common denominator t^4,

AB = √[(t^2(t+1)^2 + (1−t^2)^2)/t^4]

Expand:

t^2(t+1)^2 = t^4 + 2t^3 + t^2

(1−t^2)^2 = 1 − 2t^2 + t^4

Adding gives:

2t^4 + 2t^3 − t^2 + 1

Factorising:

2t^4 + 2t^3 − t^2 + 1 = (t+1)^2(2t^2−2t+1)

Therefore,

AB = ((t+1)√(2t^2−2t+1))/t^2

Height from O to line AB:

Using

Area = (1/2)(base)(height)

h = 2(Area)/AB

Substitute the expressions:

h = [2((1+t)/(2t^2))] / [((t+1)√(2t^2−2t+1))/t^2]

Simplifying,

h = 1/√(2t^2−2t+1)

Therefore,

Height = 1/√(2t^2−2t+1)

Finding the value of t for which the height is largest:

Since

h = 1/√(2t^2−2t+1)

the height is largest when the denominator is smallest.

Minimise:

2t^2−2t+1

Complete the square:

2t^2−2t+1

= 2(t^2−t) + 1

= 2[(t−1/2)^2 − 1/4] + 1

= 2(t−1/2)^2 + 1/2

This expression is smallest when

(t−1/2)^2 = 0

So,

t = 1/2

Therefore, the height is largest when t = 1/2.