A) \[3a\]
B) \[4a\]
C) \[5a\]
D) \[a\]
Correct Answer: D
Solution :
Let the coordinates of the point\[P\]be\[({{x}_{1}},\,\,{{y}_{1}})\]. This point lies on the curve \[{{x}^{2/3}}+{{y}^{2/3}}={{a}^{2/3}}\] ? (i) \[x_{1}^{2/3}+y_{1}^{2/3}={{a}^{2/3}}\] ? (ii) On differentiating Eq. (i) w.r.t. x, we get \[\frac{2}{3}{{x}^{-1/3}}+\frac{2}{3}{{y}^{-1/3}}\frac{dy}{dx}=0\] \[\Rightarrow \] \[\frac{dy}{dx}=-\frac{{{y}^{1/3}}}{{{x}^{1/3}}}\] \[\Rightarrow \] \[{{\left( \frac{dy}{dx} \right)}_{({{x}_{1}},\,\,{{y}_{1}})}}=-{{\left( \frac{{{y}_{1}}}{{{x}_{1}}} \right)}^{1/3}}\] The equation of the tangent at\[({{x}_{1}},\,\,{{y}_{1}})\]to the given curve is \[y-{{y}_{1}}=-\frac{y_{1}^{1/3}}{x_{1}^{1/3}}(x-{{x}_{1}})\] \[\Rightarrow \] \[xx_{1}^{-1/3}+yy_{1}^{-1/3}=x_{1}^{2/3}+y_{1}^{2/3}\] \[\Rightarrow \] \[xx_{1}^{-1/3}+yy_{1}^{-1/3}={{a}^{2/3}}\] [using Eq. (ii)] This tangent meets the coordinate axes at\[A({{a}^{2/3}},\,\,x_{1}^{1/3},\,\,0)\]and\[B(0,\,\,{{a}^{2/3}}y_{1}^{1/3})\]. \[\therefore \] \[AB=\sqrt{{{(0-{{a}^{2/3}}x_{1}^{1/3})}^{2}}+{{({{a}^{2/3}}y_{1}^{1/3}-0)}^{2}}}\] \[=\sqrt{{{a}^{4/3}}(x_{1}^{2/3}+y_{1}^{2/3})}\][from Eq. (ii)] \[=\sqrt{{{a}^{4/3}}.{{a}^{2/3}}}\] \[=\sqrt{{{a}^{2}}}=a\]
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