let (h,k) be a fixed point where h,k>0.a straight line passing through this point cuts the positive direction of coordinate axes at points P and Q.find the minimum area of the triangle OPQ

Hi Rahul,

lets take the equation of line to be x/a + y/b = 1 where line intersects x axia at ( a, 0 ) and y axis at (0,b)

Since line passes thru (h,k) therefore h/a + k/b = 1. ------- (1)

A = area of triangle OPQ = 1/2 * a*b

from eqn (1)

A = 1/2 * b * hb / ( b- K ) ( putting value of a from (1) into A )

 = 1/2*b²h/(b - k)

to calculate min value of A, dA/db = 1/2 * h * ( b² - 2bk ) / ( b-k )²

dA / db = 0 when b² - 2bk = 0

or b -2k = 0

or b = 2k

At this value of b = 2k , the d²A / db² is positive that is Minimum of A occurs at this value.

Putting in (1) a = 2h

Thus Minimum value of A occurs when a = 2h and b = 2k

Hence Min(A) = 1/2 * a * b = 1/2 * 2h * 2k = 2hk

Thus Minimum area of traingle OPQ is 2hk.

Please feel free to ask as many questions you have.

Puneet