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 equation of any line through the poit `(h, k )` is
`" " y - k = m (x - h)" " ` … (i)
For this line to intersect the positive direction of two axes, `m = tan theta lt 0`, since the angle in anti - clockwise direction from X - axis becomes obtuse.
` (##41Y_SP_MATH_C10_E04_049_S01.png" width="80%">
The line (i) meets X - axis at ` P (h - (k )/(m), 0)` and Y - axis at `Q (0, k - mh)`.
Let ` A ` = area of `Delta OPQ = (1)/(2) O P * OQ`
` " " = (1)/(2) ( h - (k )/(m)) ( k - mh )`
`" " = (1)/(2) ( (m h - k )/( m ) ) (k - mh)= - (1)/( 2m ) (k - m h)^(2)`
` " " = - (1)/( 2 tan theta ) (k - h tan theta ) ^(2) " " [because m = tan theta ]`
` " " - (1 )/( 2 tan theta ) ( k^(2)+ h^(2) tan^(2) theta - 2hk tan theta )`
` " " = (1)/(2) ( 2kh - k ^(2) cot theta - h ^(2) tan theta )`
`rArr (d A) /( dtheta ) = (1)/(2) [ - k ^(2) (-cosec^(2) theta ) -h ^(2) sec^(2) theta ]`
`" " = (1)/(2) [ k ^(2) cosec ^(2) theta - h ^(2) sec ^(2) theta] `
To obtain minimum value of `A, (dA ) /( d theta ) = 0`
`rArr k ^(2) cosec ^(2) theta - h ^(2) sec ^(2) theta = 0`
`rArr " " ( k ^(2))/( sin ^(2) theta ) = (h^(2))/( cos^(2) theta) rArr ( k ^(2))/( h ^(2)) = tan ^(2) theta `
` rArr " " tan theta = pm (k )/(h)`
` because " " tan theta lt 0, k gt 0, h gt 0" " ` [ given ]
Therefore, `tan theta = - ( k )/(h) ` ( only possible value)
Now, ` (d^(2)A)/( dtheta^(2)) = (1)/(2) [- 2k ^(2) cosec ^(2) theta cot theta - 2h ^(2) sec ^(2) theta tan theta ]`
`" " = - [ k ^(2) (1 + cot ^(2) theta ) cot theta + h ^(2) ( 1+ tan ^(2) theta) tan theta ]`
`rArr " " [ (d^(2)A)/( d theta ^(2)) ] _(tan theta = - klh)= - [ k ^(2) ( 1+ (h^(2))/( k ^(2)) ) ((-h)/(k)) + h ^(2) ( + (k ^(2))/(h^(2))) ((-k )/( h))] `
`" " =[ k ^(2) (( k ^(2) + h ^(2))/(k ^(2)) ) ((h)/(k)) + h ^(2) ((h ^(2) + k ^(2))/( h ^(2))) ((h)/(k )) ]`
`" " = [ ((k ^(2) + h ^(2) ) h) /(k ) + (( h ^(2) + k ^(2)) (k ))/( h )]`
`" " = ( k ^(2) + h^(2)) [ ( h )/(k )+(k )/(h) ] gt 0 " " [ because h, k gt 0]`
Therefore, A is least when ` tan theta =-k //h`. Also, the least value of A is
`" "A = (1)/(2) [ 2 hk -k ^(2) ((- h)/(k )) - h ^(2) ((-k )/(h))] `
` = (1)/(2) [ 2 hk + kh + kh] = 2hk`