Let `(h , k)` be a fixed point, where `h >0,k > 0.` A straight line passing through this point cuts the positive direction of the coordinate axes at the point `Pa n dQ` . Find the minimum area of triangle `O P Q ,O` being the origin.

Answer

2hk

Explanation

Let the line in intercept form be `(x)/(a)+(k)/(b)=1`
It passes through (h,k) Then `(h)/(a)+(k)/(b)=1`
or `(k)/(b)=1-(h)/(a)=(a-h)/(a)or b=(ak)/(a-h)`
`triangle` is minimum when `y =(a-h)/(a^(2))=(1)/(a)-(h)/(a^(2))` is maximum thus.
`(dy)/(da)=-(1)/(a^(2))+(2h)/(a^(3))=0 or a = 2h`
`(d^(2)y)/(da^(2))=(2)/(a^(3))=(6h)/(a^(4))=(2)/(A^(3))-(3)/(a^(3))`
or `(d^(2)y)/(da^(2))=(1)/(a^(3))= -ve`
Therefore y is maximum
Now put a =2h (1) Then `triangle =1/24h^(2)(k)/(h)=2hk`