A) \[x+y-3=0\]
B) \[x+y-1=0\]
C) \[x+y+1=0\]
D) \[x+y+3=0\]
Correct Answer: A
Solution :
\[x={{t}^{2}}\]and \[y=2t\] At \[t=1,\ x=1\]and \[y=2\] Now \[\left( \frac{dy}{dx} \right)=\frac{dy/dt}{dx/dt}=\frac{2}{2t}=\frac{1}{t}\]\[\Rightarrow \]\[{{\left( \frac{dy}{dx} \right)}_{t=1}}=1\] \[\therefore \]Equation of the normal at (1, 2) is \[y-2=-\frac{1}{\frac{dy}{dx}}(x-1)\] Þ \[y-2=-1(x-1)\]Þ \[x+y-3=0\].You need to login to perform this action.
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