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 normal at\[\left( ct,\frac{c}{t} \right)\]on the curve \[xy={{c}^{2}}\]is \[ty={{t}^{3}}x-c{{t}^{4}}+c\] If it passes through\[\left( ct,\frac{c}{t} \right)\]then \[t\frac{c}{t}={{t}^{3}}.ct-c{{t}^{4}}+c\] \[\Rightarrow \] \[t={{t}^{3}}t{{}^{2}}-{{t}^{4}}t+t\] \[\Rightarrow \] \[t-t={{t}^{3}}t(t-t)\] \[\Rightarrow \] \[1=-{{t}^{3}}t\] \[\Rightarrow \] \[t=-\frac{1}{{{t}^{3}}}\]
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