$(a)\;\frac{c^2}{a^2+b^2+c^2} \\ (b)\;\frac{c}{a+b} \\ (c)\;\frac{c^2}{a+b} \\ (d)\;None $

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From $\large\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ we have

$\large\frac{1}{b}+\frac{1}{a}=\frac{1}{f}$

or $ f= \large\frac{ab}{a+b}$----(i)

Further $AC^2+BC^2=AB^2$

or $(a^2+c^2) +(b^2+c^2)=(a+b)^2$

or $a^2+b^2+2c^2=a^2+b^2+2ab$

$\therefore ab=c^2$

Substituting this in equ (i) we gey,

$f= \large\frac{c^2}{a+b}$

Hence c is the correct answer.

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