A) \[xx_{1}^{1/3}+yy_{1}^{1/3}={{a}^{2/3}}\]
B) \[\frac{x}{{{x}^{1/3}}}+\frac{y}{{{y}^{1/3}}}={{a}^{2/3}}\]
C) \[x_{1}^{2/3}+yy_{1}^{2/3}={{a}^{1/3}}\]
D) None of the above
Correct Answer: B
Solution :
\[{{x}^{2/3}}+{{y}^{2/3}}={{a}^{2/3}}\] On differentiating with respect to\[x,\] \[\frac{2}{3}{{x}^{-1/3}}+\frac{2}{3}{{y}^{-1/3}}\frac{dy}{dx}=0\] \[\Rightarrow \] \[\frac{dy}{dx}=-\frac{{{x}^{-1/3}}}{{{y}^{-1/3}}}\] \[\Rightarrow \] \[{{\left( \frac{dy}{dx} \right)}_{({{x}_{1}},{{y}_{1}})}}=-{{\left( \frac{{{y}_{1}}}{{{x}_{1}}} \right)}^{1/3}}\] \[\therefore \]Equation of tangent \[y-{{y}_{1}}=-{{\left( \frac{{{y}_{1}}}{{{x}_{1}}} \right)}^{1/3}}(x-{{x}_{1}})\] \[\Rightarrow \] \[x_{1}^{1/3}(y-{{y}_{1}})=-y_{1}^{1/3}(x-{{x}_{1}})\] \[\Rightarrow \] \[yx_{1}^{1/3}-{{y}_{1}}x_{1}^{1/3}=xy_{1}^{1/3}-{{x}_{1}}y_{1}^{1/3}\] \[\Rightarrow \] \[xy_{1}^{1/3}+yx_{1}^{1/3}={{x}_{1}}y_{1}^{1/3}+{{y}_{1}}x_{1}^{1/3}\] On dividing both sides by\[x_{1}^{1/3}y_{1}^{1/3}\] \[\frac{x}{x_{1}^{1/3}}+\frac{y}{y_{1}^{1/3}}=\frac{{{x}_{1}}}{x_{1}^{1/3}}+\frac{{{y}_{1}}}{y_{1}^{1/3}}\] \[\Rightarrow \] \[\frac{x}{x_{1}^{1/3}}+\frac{y}{y_{1}^{1/3}}=x_{1}^{2/3}+y_{1}^{2/3}\] \[\Rightarrow \] \[\frac{x}{x_{1}^{1/3}}+\frac{y}{y_{1}^{1/3}}={{a}^{2/3}}\]
You need to login to perform this action.
You will be redirected in
3 sec
