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+1 vote
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in Mathematics by (100 points)
Find the minimum length of hypotenuse of a triangle if a point P is on hypotenuse so that it is at a distance 'a' and 'b' from the sides of the triangle

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Best answer

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

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