A) only circles
B) only rectangular hyperbolas
C) either circles or rectangular hyperbolas
D) None of the above
Correct Answer: C
Solution :
We have length of the normal = radius vector \[\Rightarrow \] \[y\sqrt{1+{{\left( \frac{dy}{dx} \right)}^{2}}}=\sqrt{{{x}^{2}}+{{y}^{2}}}\] \[\Rightarrow \] \[{{y}^{2}}\left\{ 1+{{\left( \frac{dy}{dx} \right)}^{2}} \right\}={{x}^{2}}+{{y}^{2}}\] \[\Rightarrow \] \[{{y}^{2}}{{\left( \frac{dy}{dx} \right)}^{2}}={{x}^{2}}\] \[\Rightarrow \] \[x=\pm y\frac{dy}{dx}\] \[\Rightarrow \] \[x=y\frac{dy}{dx}\] or \[x=-y\frac{dy}{dx}\] \[\Rightarrow \] \[x\,\,dx-y\,\,dy=0\] or \[x\,\,dx+y\,\,dy=0\] \[\Rightarrow \] \[{{x}^{2}}-{{y}^{2}}={{c}_{1}}\] or \[{{x}^{2}}+{{y}^{2}}={{c}_{2}}\] Clearly,\[{{x}^{2}}-{{y}^{2}}={{c}_{1}}\]represents a rectangular hyperbola and\[{{x}^{2}}+{{y}^{2}}={{c}_{2}}\]represents circle.
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