A) \[a+b=c\]
B) \[a-b=c\]
C) \[a+2b=c\]
D) \[2a-b=c\]
Correct Answer: A
Solution :
\[{{x}^{\frac{2}{3}}}\,+{{y}^{\frac{2}{3}}}={{c}^{\frac{2}{3}}}\,\,\,\,\,\,\,\,\,\,;\,\,\,\,\,\,\,\,\,\,\,\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] \[\Rightarrow \,\frac{2}{3}\,{{x}^{\frac{-1}{3}}}\,+\frac{2}{3}\,{{y}^{\frac{-1}{3}}}\,\frac{dy}{dx}=0;\,\,\frac{x{{x}_{1}}}{{{a}^{2}}}\,+\frac{y{{y}_{1}}}{{{b}^{2}}}=1\] \[\Rightarrow \,\frac{dy}{dx}=\frac{-y_{1}^{1/3}}{x_{1}^{1/3}}\] \[\therefore \,\,(y-{{y}_{1}})=\left( \frac{-y_{1}^{1/3}}{x_{1}^{1/3}} \right)\,(x-{{x}_{1}})\] \[yx_{1}^{1/3}\,-{{y}_{1}}x_{1}^{1/3}\,=-y_{1}^{1/3}\,x+{{x}_{1}}y_{1}^{1/3}\] \[y_{1}^{1/3}\,x+x_{1}^{1/3}\,y=x_{1}^{1/3}\,{{y}^{1/3}}\,(x_{1}^{1/3}\,+y_{1}^{2/3})\] \[\left. \begin{align} & \frac{x}{x_{1}^{1/3}}+\frac{y}{y_{1}^{1/3}}\,={{c}^{2/3}} \\ & \frac{x{{x}_{1}}}{{{a}^{2}}}\,+\frac{y{{y}_{1}}}{{{b}^{2}}}=1 \\ \end{align} \right]\,\] identical \[\frac{a}{{{c}^{1/3}}}\,+\frac{b}{{{c}^{1/3}}}={{c}^{2/3}}\] \[\frac{x_{1}^{4/3}\,}{{{a}^{2}}}\,=\frac{y_{1}^{4/3}}{{{b}^{2}}}\,=\frac{1}{{{c}^{2/3}}}\] \[\left. \begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,x_{1}^{2/3}\,=\frac{a}{{{c}^{1/3}}} \\ & |||1y\,\,\,\,\,\,y_{1}^{2/3}=\frac{b}{{{c}^{1/3}}} \\ \end{align} \right]\Rightarrow \,x_{1}^{2/3}\,+y_{1}^{2/3}=\,\frac{a+b}{{{c}^{1/3}}}\] \[\Rightarrow \,\,{{c}^{2/3}}\,=\frac{a+b}{{{c}^{1/3}}}\,=a+b=c\]
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