Question

If a point on the hypotenuse of a triangle is at distance $a$ and $b$ from the sides of triangle, then the minimum length of the hypotenuse is

• A. $\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)$
• B. ${\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)}^{\frac{3}{2}}$
• C. ${\left(a^{\frac{1}{3}}+b^{\frac{1}{3}}\right)}^{\frac{3}{2}}$
• D. None of the above

Answer 1

Answer: (B)

Let $P$ be a point on the hypotenuse $AC$ of right angled $△ABC$.
Such that $PL\perp AB=a$ and $PM\perp BC=b$
Hence $PL=a$
$PM=b$
Clearly, $\angle APL=\angle ACB=\theta$ (say)
$AP=a\sec \theta ,PC=b\mathrm{cosec}\theta$
Let $l$ be the length of the hypotenuse, then
$l=AP+PC$
$\Rightarrow l=a\sec \theta +b\mathrm{cosec}\theta ,0<\theta <\frac{\pi }{2}$
On ditterentiating w.r.t. $\theta$, we get
$\frac{dl}{d\theta }=a\sec \theta \tan \theta -b\mathrm{cosec}\theta \cot \theta$

For maxima or minima put $\frac{dl}{d\theta }=0$
$\Rightarrow a\sec \theta \tan \theta =b\mathrm{cosec}\theta \cot \theta$
$\Rightarrow \frac{a\sin \theta }{{\cos }^2\theta }=\frac{b\cos \theta }{{\sin }^2\theta }$
$\Rightarrow \frac{{\sin }^3\theta }{{\cos }^3\theta }=\frac{b}{a}$
$\Rightarrow {\tan }^3\theta =\frac{b}{a}$
$\Rightarrow \tan \theta ={\left(\frac{b}{a}\right)}^{1/3}$
Now, $\frac{d^{2}l}{d{\theta }^2}=a\left(\sec \theta \times {\sec }^2\theta +\tan \theta \times \sec \theta \tan \theta \right]$
$=a\mathrm{cosec}\theta \left(-{\mathrm{cosec}}^2\theta \right)+\cot \theta \left(-{\mathrm{cosec}}^2\theta {\cot }^2\theta \right)]$
$=a\left(\sec \theta +{\tan }^2\theta \right)+b\mathrm{cosec}\theta \left(\mathrm{cosec}{}^2\theta +{\cot }^2\theta \right)$
Since, $0<\theta <\frac{\pi }{2}$, so trigonometric ratios are positive.
Also, $a>0$ and $b>0$.
$\therefore \frac{d^{2}I}{d{\theta }^2}$ is positive.
$\Rightarrow l$ is least when $\tan \theta ={\left(\frac{b}{a}\right)}^{1/3}$
Now, we have the following figure

$\therefore$ Least value of $I=a\sec \theta +b\mathrm{cosec}\theta$
$=a\frac{\sqrt{a^{2/3}+b^{2/3}}}{a^{1/3}}+b\frac{\sqrt{a^{2/3}+b^{2/3}}}{b^{1/3}}$
$=\sqrt{a^{2/3}+b^{2/3}}\left(a^{2/3}+b^{2/3}\right)={\left(a^{2/3}+b^{2/3}\right)}^{3/2}$
$\left[\because \text{ in }\Delta EFG,\sec \theta =\frac{\sqrt{a^{2/3}+b^{2/3}}}{a^{1/3}}\text{ and }\mathrm{cosec}\theta =\frac{\sqrt{a^{2/3}+b^{2/3}}}{b^{1/3}}\right]$