A) \[x+9y-6=0\]
B) \[9x-y-6=0\]
C) \[9x+y+6=0\]
D) \[x+9y+6=0\]
E) \[9x+y-6=0\]
Correct Answer: E
Solution :
Curve, \[x=\frac{t-1}{t+1},y=\frac{t+1}{t-1}\] At\[t=2,\text{ }x=\frac{1}{3}\]and\[y=3\] Now, \[\frac{dx}{dt}=\frac{(t+1).1-(t-1).1}{{{(t+1)}^{2}}}=\frac{2}{{{(t+1)}^{2}}}\] \[\frac{dy}{dx}=\frac{(t-1).1-(t+1).1}{{{(t-1)}^{2}}}=\frac{-2}{{{(t-1)}^{2}}}\] Now, \[\frac{dy}{dx}=\frac{dy}{dx}.\frac{dt}{dx}=\frac{-2}{{{(t-1)}^{2}}}.\frac{{{(t+1)}^{2}}}{2}\] \[\frac{dy}{dx}=-{{\left( \frac{t+1}{t-1} \right)}^{2}}\] \[{{\left( \frac{dy}{dx} \right)}_{at(t=2)}}=-{{\left( \frac{3}{1} \right)}^{2}}=-9\] So, the equation of line is \[(y-3)=-9(x-1/3)\] \[(y-3)=-3(3x-1)\] \[y-3=-9x+3\] \[\Rightarrow \] \[9x+y-6=0\]
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