A) 0
B) \[\frac{1}{2}\]
C) \[\infty \]
D) \[-2\]
Correct Answer: A
Solution :
\[x=3{{t}^{2}}+1,\,y={{t}^{3}}-1\] \ \[\frac{dx}{dt}=6t,\] \[\frac{dy}{dt}=3{{t}^{2}}\] Now \[\frac{dy}{dx}=\left( \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \right)\] = \[\frac{3{{t}^{2}}}{6t}=\frac{t}{2}\] For \[x=1\], \[3{{t}^{2}}+1=1\Rightarrow t=0\] Þ Slope = \[\frac{0}{2}=0\].You need to login to perform this action.
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