Let P be a point on the hypotenuse AC of right △ABC such that
PL⊥AB=a
PM⊥BC=b
Let ∠APL=∠ACB=θ
AP=asecθ and PC=bcosecθ
Let l be the length of the hypotenuse, then l=AP+PC
⇒ asecθ + bcosecθ , (0< θ < pi/2)
Now differentiate l with respect to θ, we get
dθ/dl=asecθ.tanθ−bcosecθ.cotθ
For maxima and minima dθ/dl=0
Thus asecθ.tanθ=bcosecθ.cotθ
⇒asin3θ=bcos3θ
b/a=sin3θ/cos3θ
Thus b/a=tan3θ
⇒tanθ=[b/a]1/3
Now dθ2/d2l=a(secθ.sec2θ+tanθ.secθ.tanθ)
=−b(cosecθ(−cosec2θ) + cotθ(−cosecθ.cotθ)
=asecθ(sec2θ+tan2θ) + bcosecθ × bcosecθ + cot2θ
Since 0<θ< pi/2 , so all t ratios of θ are positive.
Also a>0 and b<0
Therefore, dθ2/ d2l is positive.
⇒ l is least when tanθ=(b/a)1/3
Least value of l =asecθ+bcosecθ
= a{[√(a2/3 + b2/3)]/a1/3 } + b{[√(a2/3 + b2/3)]/b1/3 }
= [√(a2/3 + b2/3)] * (a2/3 + b2/3)
= (a2/3 + b2/3)3/2
.