N}}{{x}^{n}}\] or \[\frac{dy}{dx}={{a}^{1-n}}n{{x}^{n-1}}\] \[sub-normal=\LEFT| y\frac{dy}{dx} \right|=\left| y{{a}^{1-n}}n{{x}^{n-1}} \right|\] \[=\left| {{a}^{1-n}}{{x}^{n}}{{a}^{1-n}}n{{x}^{n-1}} \right|=\left| {{a}^{2-2n}}{{x}^{2n-1}} \right|\] which is CONSTANT if \[2n-1=0\] or \[n=1/2\].

"> N}}{{x}^{n}}\] or \[\frac{dy}{dx}={{a}^{1-n}}n{{x}^{n-1}}\] \[sub-normal=\LEFT| y\frac{dy}{dx} \right|=\left| y{{a}^{1-n}}n{{x}^{n-1}} \right|\] \[=\left| {{a}^{1-n}}{{x}^{n}}{{a}^{1-n}}n{{x}^{n-1}} \right|=\left| {{a}^{2-2n}}{{x}^{2n-1}} \right|\] which is CONSTANT if \[2n-1=0\] or \[n=1/2\].

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If the sub-normal at any point on \[y={{a}^{1-n}}{{x}^{n}}\] is of constant length, then the value of n is

Joint Entrance Exam - Main (JEE Main) Mathematics in Joint Entrance Exam - Main (JEE Main) . 10 months ago

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[d] \[y={{a}^{1-N}}{{x}^{n}}\] or \[\frac{dy}{dx}={{a}^{1-n}}n{{x}^{n-1}}\] \[sub-normal=\LEFT| y\frac{dy}{dx} \right|=\left| y{{a}^{1-n}}n{{x}^{n-1}} \right|\] \[=\left| {{a}^{1-n}}{{x}^{n}}{{a}^{1-n}}n{{x}^{n-1}} \right|=\left| {{a}^{2-2n}}{{x}^{2n-1}} \right|\] which is CONSTANT if \[2n-1=0\] or \[n=1/2\].

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Posted on 14 Aug 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Mathematics in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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