A) \[{{c}^{-4/3}}\]
B) \[{{c}^{-1/2}}\]
C) \[{{c}^{1/2}}\]
D) None of these
Correct Answer: A
Solution :
Let \[({{x}_{1}},{{y}_{1}})\] be a point on the given curve. We have, \[{{x}^{4}}+{{y}^{4}}={{c}^{4}}\] \[\Rightarrow \]\[4{{x}^{3}}+4{{y}^{3}}\frac{dy}{dx}=0\]\[\Rightarrow \]\[\frac{dy}{dx}=-\frac{{{x}^{3}}}{{{y}^{3}}}\]\[\Rightarrow \] \[\Rightarrow \]\[{{\left( \frac{dy}{dx} \right)}_{({{x}_{1}},{{y}_{1}})}}=-\frac{x_{1}^{3}}{y_{1}^{3}}\] The equation of the tangent at \[({{x}_{1}},{{y}_{1}})\] is \[y-{{y}_{1}}=-\frac{x_{1}^{3}}{y_{1}^{3}}(x-{{x}_{1}})\] \[\Rightarrow \]\[yy_{1}^{3}-y_{1}^{4}=-xx_{1}^{3}+x_{1}^{4}\] \[\Rightarrow \]\[xx_{1}^{3}+yy_{1}^{3}=x_{1}^{4}+y_{1}^{4}\] \[\Rightarrow \]\[xx_{1}^{3}+yy_{1}^{3}={{c}^{4}}\] [\[\because ({{x}_{1}},{{y}_{1}})\]lies on \[{{x}^{4}}+{{y}^{4}}={{c}^{4}}\]] The lengths of intercepts made on the coordinate axes by this tangent are \[\frac{{{c}^{4}}}{x_{1}^{3}}\] and \[\frac{{{c}^{4}}}{x_{1}^{3}}\]respectively. \[\therefore \]\[a=\frac{{{c}^{4}}}{x_{1}^{3}}\]and\[b=\frac{{{c}^{4}}}{x_{1}^{3}}\] \[\Rightarrow \]\[x_{1}^{3}=\frac{{{c}^{4}}}{a}\]and \[y_{1}^{3}=\frac{{{c}^{4}}}{b}\] \[\Rightarrow \]\[{{x}_{1}}={{\left( \frac{{{c}^{4}}}{a} \right)}^{1/3}}\]and \[{{y}_{1}}={{\left( \frac{{{c}^{4}}}{b} \right)}^{1/3}}\] Since, \[({{x}_{1}},{{y}_{1}})\]lies on\[{{x}^{4}}+{{y}^{4}}={{c}^{4}}.\]Therefore, \[x_{1}^{4}+y_{1}^{4}={{c}^{4}}\Rightarrow {{\left( \frac{{{c}^{4}}}{a} \right)}^{4/3}}+{{\left( \frac{{{c}^{4}}}{b} \right)}^{4/3}}={{c}^{4}}\] \[\Rightarrow \]\[{{a}^{-4/3}}+{{b}^{-4/3}}={{c}^{-4/3}}\]
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