Answer:
Here \[u=-(f+a),u=-(f+b),f=-f\] As \[\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\] \[\therefore \] \[f=\frac{uv}{u+v}\] or \[-f=\frac{[-(f+a)\times [-(f+b)]}{-(f+a)-f(a+b)}\] \[=\frac{{{f}^{2}}+af+bf+ab}{-(2f+a+b)}\] or \[2{{f}^{2}}+af+bf={{f}^{2}}+af+bf+ab\] or \[{{f}^{2}}=ab\]
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