A) \[t=0\]
B) \[t=\infty \]
C) \[t=\frac{1}{\sqrt{3}}\]
D) \[t=\frac{-1}{\sqrt{3}}\]
Correct Answer: A
Solution :
Given, \[x={{t}^{2}}-1\] and \[y={{t}^{2}}-t\] Now, \[\frac{dx}{dt}=2t\] and \[\frac{dy}{dt}=2t-1\] \[\therefore \] \[\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2t-1}{2t}\] \[\because \] The tangent line is perpendicular to x-axis. \[\therefore \] \[\frac{dy}{dx}=\tan \,{{90}^{o}}=\infty =\frac{1}{0}\] \[\Rightarrow \] \[\frac{2t-1}{2t}=\frac{1}{0}\] \[\Rightarrow \] \[t=0\] where, \[\left( t\ne \frac{1}{2} \right)\]
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