A) \[{{(abscissa)}^{2}}\]
B) \[{{(abscissa)}^{3}}\]
C) abscissa
D) ordinate
Correct Answer: C
Solution :
We have,\[{{x}^{m}}{{y}^{n}}={{a}^{m+n}}\] Taking logarithm on both sides, we get \[\Rightarrow m\log x+n\log y=(m+n)\log a\] Differentiating both sides w.r.t.\[x,\] we get \[\therefore \] \[\frac{m}{x}+\frac{n}{y}\frac{dy}{dx}=0\] \[\Rightarrow \] \[\frac{dx}{dy}=-\frac{nx}{my}\] \[\therefore \]Subtangent\[=\left| y\frac{dx}{dy} \right|=\frac{nx}{m}\propto x\]
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