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akshay chaturvedi

Let (h,k) be a fixed point, where h>0, 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 , O being the origin.

rini s

is the ans= 2kh

akshay chaturvedi

yes the answer is 2kh but u please give the solution

Ramya Hegde

Let the slope of the line be m.

SO the eqn of line becomes,

y-k =m(x-h)

mx-y+k-hm=0

Let P be the point on y-axis and Q on x-axis.

so,

P(O,k-hm)

Q(mh-k/m ,0)

so area of the triangle becomes,

A=1/2(mh-k/m)(k-hm)

= 1/2 ( 2hk -k²/m -h²m)

hen area is at its minimum, its derivative becomes zero.

On differenciating, the expression,

A^' = 1/2( k²m² -h²)

equate it to zero, and u get two values of m=k/h, -k/h

now for minimum value of area, its second derivative is positive.

so, A^'' = -k²/m³

and it is positive for the negative value of m.

Now, on substituting back this value of m, in the expression of area, u get,

A=1/2(2hk+kh+hk)

=2hk

Hence, the answer.

akshay chaturvedi

is there any other method to solve this question?????????

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