A) (2a, a)
B) \[(-2a,a)\]
C) (a, 2a)
D) \[(a,-2a)\]
Correct Answer: A
Solution :
Curve\[{{x}^{3}}=8{{a}^{2}}y\] Differentiating with respect to\[x,\] \[3{{x}^{2}}=8{{a}^{2}}\frac{dy}{dx}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{3{{x}^{2}}}{8{{a}^{2}}}\] Slope of normal \[=-\frac{1}{\frac{dy}{dx}}-\frac{1}{\frac{3{{x}^{2}}}{8{{a}^{2}}}}=-\frac{8{{a}^{2}}}{3{{x}^{2}}}\] But \[-\frac{8{{a}^{2}}}{3{{x}^{2}}}=-\frac{2}{3}\] \[\Rightarrow \] \[4{{a}^{2}}={{x}^{2}}\] \[\Rightarrow \] \[x=\pm 2a\] In first quadrant \[x=2a\] \[\therefore \] \[y=\frac{8{{a}^{3}}}{8{{a}^{2}}}=a\] Hence, point is\[(2a,a)\].
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