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A point on the hypotenuse of …

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A point on the hypotenuse of a right triangle is ata a distance a and b from the sides ofa triangel. Show that the minimum lenght of the hypotenuse is (a2/3+b2/3)3/2.
  • 1 answers

Sia ? 6 years, 6 months ago


{tex}AP = a\cos ec\theta {/tex} 
{tex}BP = b\sec \theta {/tex} 
{tex}l = AP + BP{/tex} 
{tex}l = a\cos ec\theta + b\sec \theta {/tex} 
{tex}\frac{{dl}}{{d\theta }} = -a\cos ec\theta \cot \theta + b\sec \theta .\tan \theta {/tex} 
{tex}\frac{{{d^2}l}}{{d{\theta ^2}}} = a\cos e{c^2}\theta + a\cos ec\theta {\cot ^2}\theta + {b^2} - {\sec ^3}\theta + b\sec \theta .{\tan ^2}\theta {/tex} 
For maximum/minimum
{tex}\frac{{dl}}{{d\theta }} = 0{/tex} 
{tex}\frac{{dl}}{{d\theta }} = -a\cos ec\theta \cot \theta + b\sec \theta .\tan \theta {/tex} 
{tex}\frac{acos\theta}{sin^2\theta}=\frac{bsin\theta}{cos^2\theta}{/tex} 
{tex}{\tan ^3}\theta = \frac{a}{b}{/tex} 
{tex}\tan \theta = {\left( {\frac{a}{b}} \right)^{1/3}}{/tex} 
{tex}\sin \theta = \frac{{{a^{1/3}}}}{{\sqrt {{a^{2/3}} + {b^{2/3}}} }},\cos \theta = \frac{{{b^{1/3}}}}{{\sqrt {{a^{2/3}} + {b^{2/3}}} }}{/tex} 
{tex}\frac{{{d^2}l}}{{d{\theta ^2}}} < 0{/tex} for {tex}\tan \theta = {\left( {\frac{a}{b}} \right)^{1/3}}{/tex} 
l  is minimum and minimum value of l is given by,
{tex}l = a\cos ec\theta + b\sec \theta {/tex} 
{tex}=a\sqrt{1+cot^2\theta}+b\sqrt{1+tan^2\theta}{/tex} 
{tex}=a\sqrt{1+(\frac{b}{a})^{2/3}}+b\sqrt{1+(\frac{a}{b})^{2/3}}{/tex} 
{tex}=a^{2/3}\sqrt{a^{2/3}+b^{2/3}}+b^{2/3}\sqrt{b^{2/3}+a^{2/3}}{/tex} 
{tex}l = {\left( {{a^{2/3}} + {b^{2/3}}} \right)^{3/2}}{/tex}

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