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 $A C$ of right angled $\triangle A B C$.
Such that $P L \perp A B=a$ and $P M \perp B C=b$
Hence $ P L=a$
$P M=b$
Clearly, $ \angle A P L=\angle A C B=\theta $ (say)
$A P=a \sec \theta, P C=b \operatorname{cosec} \theta$
Let $l$ be the length of the hypotenuse, then
$l =A P+P C $
$\Rightarrow l =a \sec \theta+b \operatorname{cosec} \theta, 0 < \theta < \frac{\pi}{2}$
On ditterentiating w.r.t. $\theta$, we get
$\frac{d l}{d \theta}=a \sec \theta \tan \theta-b \operatorname{cosec} \theta \cot \theta$

For maxima or minima put $\frac{d l}{d \theta}=0$
$\Rightarrow a \sec \theta \tan \theta=b \operatorname{cosec} \theta \cot \theta$
$\Rightarrow \frac{a \sin \theta}{\cos ^2 \theta}=\frac{b \cos \theta}{\sin ^2 \theta}$
$\Rightarrow \frac{\sin ^3 \theta}{\cos ^3 \theta}=\frac{b}{a}$
$\Rightarrow \tan ^3 \theta=\frac{b}{a}$
$\Rightarrow \tan \theta=\left(\frac{b}{a}\right)^{1 / 3}$
Now, $\frac{d^2 l}{d \theta^2}=a\left(\sec \theta \times \sec ^2 \theta+\tan \theta \times \sec \theta \tan \theta\right]$
$\left.=a \operatorname{cosec} \theta\left(-\operatorname{cosec}^2 \theta\right)+\cot \theta\left(-\operatorname{cosec}^2 \theta \cot ^2 \theta\right)\right]$
$=a\left(\sec \theta+\tan ^2 \theta\right)+b \operatorname{cosec} \theta\left(\operatorname{cosec}{ }^2 \theta+\cot ^2 \theta\right)$
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 $I=a \sec \theta+b \operatorname{cosec} \theta$
$=a \frac{\sqrt{a^{2 / 3}+b^{2 / 3}}}{a^{1 / 3}}+b \frac{\sqrt{a^{2 / 3}+b^{2 / 3}}}{b^{1 / 3}}$
$=\sqrt{a^{2 / 3}+b^{2 / 3}}\left(a^{2 / 3}+b^{2 / 3}\right)=\left(a^{2 / 3}+b^{2 / 3}\right)^{3 / 2}$
$\left[\because \text { in } \Delta E F G, \sec \theta=\frac{\sqrt{a^{2 / 3}+b^{2 / 3}}}{a^{1 / 3}} \text { and } \operatorname{cosec} \theta=\frac{\sqrt{a^{2 / 3}+b^{2 / 3}}}{b^{1 / 3}}\right]$