A) \[c=\frac{1}{\sqrt{{{\mu }_{0}}{{\varepsilon }_{0}}}}\]
B) \[I\propto \frac{1}{{{\lambda }^{4}}}\]
C) \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}={{\left[ \frac{{{I}_{2}}}{{{I}_{1}}} \right]}^{\frac{1}{4}}}={{\left( \frac{4}{1} \right)}^{\frac{1}{4}}}=\sqrt{2}:1\]
D) \[K=\frac{M}{\sqrt{4{{L}_{2}}}}\]
Correct Answer: C
Solution :
Let the ends of other axis of symmetry be\[(0,\pm a)\]. Then, the equation of the ellipse is \[{{x}^{2}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1\] \[\Rightarrow \] \[{{a}^{2}}{{x}^{2}}+{{y}^{2}}={{a}^{2}}\] ...(i) On differentiating both sides w.r.t. x, we get \[2{{a}^{2}}x+2y\frac{dy}{dx}=0\] \[\Rightarrow \] \[{{a}^{2}}=-\frac{y}{x}.\frac{dy}{dx}\] ??? (ii) On eliminating \[{{a}^{2}}\]from Eqs. (i) and (ii), we get \[-y\frac{dy}{dx}+{{y}^{2}}=-\frac{y}{x}.\frac{dy}{dx}\] \[\Rightarrow \] \[({{x}^{2}}-1)\frac{dy}{dx}-xy=0\]
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