A) \[a=3,b=-1\]
B) \[a=3,b=1\]
C) \[a=-3,b=1\]
D) \[a=-3,b=-1\]
Correct Answer: A
Solution :
We have, \[x={{t}^{2}}+3t-8\] \[\Rightarrow \] \[\frac{dx}{dt}=2t+3\] and \[y=2{{t}^{2}}-2t-5\] \[\Rightarrow \] \[\frac{dy}{dt}=4t-2\] Now, \[\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{4t-2}{2t+3}\] Now, for\[x=2\]and\[y=-1,\]we have \[{{t}^{2}}+3t-8=2\]and \[2{{t}^{2}}-2t-5=-1\] \[\Rightarrow \] \[{{t}^{2}}+3t-10=0\] and \[{{t}^{2}}-t-2=0\] Solving these two equations, we get \[t=2\] \[\therefore \] \[{{\left( \frac{dy}{dx} \right)}_{(2,-1)}}=\frac{4\times 2-2}{2\times 2+3}=\frac{6}{7}\]
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