question_answer

A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Then the minimum length of the hypotenuse is

A) \[{{\left( {{a}^{\frac{3}{2}}}+{{b}^{\frac{3}{2}}} \right)}^{\frac{2}{3}}}\]

B) \[{{\left( {{a}^{\frac{2}{3}}}+{{b}^{\frac{2}{3}}} \right)}^{\frac{3}{2}}}\]

C) \[{{\left( {{a}^{\frac{2}{3}}}+{{b}^{\frac{2}{3}}} \right)}^{3}}\]

D) \[{{\left( {{a}^{\frac{3}{2}}}+{{b}^{\frac{3}{2}}} \right)}^{3}}\]

Correct Answer: B

Solution :

[b]

From the figure,

\[PC=b\,\,\cos ec\theta \]  and \[AP=a\,\,sec\theta \]

\[AC=PC+AP\]

or \[AC=b\,\,\cos ec\theta +a\,\,\sec \theta ....(1)\]

\[\therefore \frac{d(AC)}{d\theta }=-b\,\,\cos ec\theta \cot \theta +a\,\,\sec \theta \tan \theta \]

For minimum length, \[\frac{d(AC)}{d\theta }=0\]

or

\[a\sec \theta \tan \theta =b\cos ec\theta \cot \theta \] or \[\tan \theta ={{\left( \frac{b}{a} \right)}^{1/3}}\]

\[\therefore \sin \theta =\frac{{{(b)}^{1/3}}}{\sqrt{{{a}^{2/3}}+{{b}^{2/3}}}}\] and

\[\cos \theta =\frac{{{(a)}^{1/3}}}{\sqrt{{{a}^{2/3}}+{{b}^{2/3}}}}\]

Also, \[\theta \in (0,\pi /2)\]

\[\underset{\theta \to 0}{\mathop{\lim }}\,(a\,\sec \theta +b\,\cos ec\theta )\to \infty \]

and \[\underset{\theta \to \pi /2}{\mathop{\lim }}\,(a\,\sec \theta +b\,\cos ec\theta )\to \infty \]

Therefore, \[\theta ={{\tan }^{-1}}{{\left( \frac{b}{a} \right)}^{1/3}}\] is a point of minima.

For this value of \[\theta \],

\[AC=\frac{b\sqrt{{{a}^{2/3}}+{{b}^{2/3}}}}{{{b}^{1/3}}}+\frac{a\sqrt{{{a}^{2/3}}+{{b}^{2/3}}}}{{{a}^{1/3}}}\]

[Using (1) and (2)]

\[=\sqrt{{{a}^{2/3}}+{{b}^{2/3}}}({{b}^{2/3}}+{{a}^{2/3}})={{({{a}^{2/3}}+{{b}^{2/3}})}^{3/2}}\]

Hence, the minimum length of the hypotenuse is

\[{{({{a}^{2/3}}+{{b}^{2/3}})}^{3/2}}.\]