A) \[t'=\frac{-1}{{{t}^{3}}}\]
B) \[t'=\frac{-1}{t}\]
C) \[t'=\frac{1}{{{t}^{2}}}\]
D) \[t{{'}^{2}}=\frac{-1}{{{t}^{2}}}\]
Correct Answer: A
Solution :
The equation of the tangent at\[\left( ct,\,\,\frac{c}{t} \right)\]is \[ty={{t}^{3}}x-c{{t}^{4}}+c\] If it passes through\[\left( ct',\,\,\frac{c}{t'} \right)\], then \[\frac{tc}{t'}={{t}^{3}}ct'-c{{t}^{4}}+c\] \[\Rightarrow \] \[t={{t}^{3}}t{{'}^{2}}-{{t}^{4}}t'+t'\] \[\Rightarrow \] \[t-t'={{t}^{3}}t'(t'-t)\] \[\Rightarrow \] \[t'=\frac{1}{{{t}^{3}}}\]
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