Question

A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the minimum length of the hypotenuse is $(a^{\frac{3}{2}}+b^{\frac{2}{3}}{)}^{\frac{3}{2}}$

Answer 1

Let P be a point on the hypotenuse AC of right angled $\Delta ABC.$ Such that $PL\perp AB=aandPM\perp BC=b$
Let $\angle APL=\angle ACB=\theta$ (say)
$AP=asec\theta ,PC=bcosec\theta$
Let l be the length of the hypotenuse, then
$AP=asec\theta ,PC=bcosec\theta$
Let l be the length of the hypotenuse, then
$l=AP+PC\Rightarrow l=asec\theta +bcosec\theta ,0<\theta <\frac{\pi }{2}$
On differentiating w.r.t. $\theta ,$ we get
$\frac{dl}{d\theta }=asec\theta tan\theta -bcosec\theta cot\theta$
For maxima or minima put $\frac{dl}{d\theta }=0$
$\Rightarrow asec\theta =bcosec\theta cot\theta \Rightarrow \frac{asin\theta }{co{s}^2\theta }=\frac{bcos\theta }{si{n}^2\theta }\Rightarrow tan\theta =(\frac{b}{a}{)}^{\frac{1}{3}}$

Now, $\frac{d^{2}l}{d{\theta }^2}=a(sec\theta \times se{c}^2\theta +tan\theta \times sec\theta tan\theta ]$
$-b\left[cosec\theta \right(-cose{c}^2\theta )+cot\theta (-cosec\theta cot\theta \left)\right]=asec\theta (se{c}^2\theta +ta{n}^2\theta )+bcosec\theta (cose{c}^2\theta +co{t}^2\theta )$
Since, $0<\theta <\frac{\pi }{2},$ so, trigonometric ratios are positve.
Also, a > 0 and b > 0.
$\therefore \frac{d^{2}l}{d{\theta }^2}$ is positive.
$\Rightarrow$ l is least when $tan\theta =(\frac{b}{a}{)}^{\frac{1}{3}}$
$\therefore$ Least value of $l=asec\theta +bcosec\theta$
$=a\frac{\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}{a^{\frac{1}{3}}}+b=a\frac{\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}{a^{\frac{1}{3}}}=\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}(a^{\frac{2}{3}}+b^{\frac{2}{3}})=(a^{\frac{2}{3}}+b^{\frac{2}{3}}{)}^{\frac{3}{2}}$
$\because In\Delta EFG,sec\theta =\frac{\sqrt{a^{\frac{2}{2}}+b^{\frac{2}{3}}}}{a^{\frac{1}{3}}}andcosec\theta =\frac{\sqrt{a^{\frac{2}{2}}+b^{\frac{2}{3}}}}{a^{\frac{1}{3}}}$
Hence proved.