Question

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

• A. $(a + b)^{3/2}$
• B. $a^{2/3} + b^{2/3}$
• C. $(a^{2/3} + b^{2/3})^2$
• D. $(a^{2/3} + b^{2/3})^{3/2}$

Answer 1

Answer: (D)

Let $AOB$ be a right triangle with hypotenuse $AB$ such that a point $P$ on $AB$ is at distances $a$ and $b$ from $OA$ and $OB$ respectively, i.e. $PL = a$ and $PM = b$
Let $∠OAB = \theta$. Then,
$AP - a \,cosec \,\theta$ and $BP = b \,sec\, \theta$
Let $l$ be the length of the hypotenuse $AB$. Then,
$l = AP + BP$
$\Rightarrow l = a \,cosec \,\theta + b \,sec\, \theta $
$\Rightarrow \frac{dl}{d\theta} = - a \,cosec \,\theta \,cot\,\theta + b \,sec\, \theta \,tan\,\theta$
and, $\frac{d^{2}l}{d\theta^{2} } = a \,cosec^{3} \,\theta +a \,cosec\, \theta \,cot^{2}\,\theta + b \,sec^{3}\, \theta + b \,sec\, \theta \,tan^{2}\,\theta $
For maximum or minimum, we must have
$\frac{dl}{d\theta } = 0$
$\Rightarrow - a \,cosec \,\theta \,cot\,\theta + b \,sec\, \theta$
$tan\,\theta = 0$
$\Rightarrow -\frac{a \,cos \,\theta }{sin^{2}\,\theta} + \frac{b\,sin\,\theta}{cos^{2}\,\theta } = 0$

$\Rightarrow tan^{3}\,\theta = \frac{a}{b}$
$\Rightarrow tan\,\theta = \left(\frac{a}{b}\right)^{1/3}$
$\Rightarrow sin\,\theta =\frac{a^{1/3}}{\sqrt{a^{2/3}+ b^{2/3}}}$ and
$cos \,\theta = \frac{b^{1/3}}{\sqrt{a^{2/3}+ b^{2/3}}}$
Clearly, $\frac{d^{2}l}{d\theta ^{2} } > 0$ for $tan\,\theta = \left(\frac{a}{b}\right)^{1/3}$
Thus, $l$ is minimum when $tan\,\theta = \left(\frac{a}{b}\right)^{1/3}$
The minimum value of $l$ is given by
$l = a \,cosec \,\theta + b \,sec\, \theta$
$=a\sqrt{1+cot^{2}\,\theta } + b\sqrt{1+tan^{2}\,\theta }$
$l = a\sqrt{1+\left(\frac{b}{a}\right)^{2/3}} +b\sqrt{1+\left(\frac{a}{b}\right)^{2/3}}$
$\Rightarrow l = (a^{2/3} + b^{2/3})^{3/2}$