A) \[3x-y-2=0\]
B) \[2x+3y-7=0\]
C) \[2x-3y+1=0\]
D) \[2y-3x+1=0\]
Correct Answer: B
Solution :
Given, \[{{y}^{2}}={{x}^{3}}\] On differentiating w.r.t.\[x,\]we get \[2y\frac{dy}{dx}=3{{x}^{2}}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{3{{x}^{2}}}{2y}\] \[\Rightarrow \] \[{{\left( \frac{dy}{dx} \right)}_{(1,1)}}=\frac{3}{2}\] \[\therefore \]Equation of normal at point (1, 1) is \[y-1=-\frac{2}{3}(x-1)\] \[\Rightarrow \] \[2x+3y=5\] Hence, the equation of line parallel to above line will be \[2x+3y=7\]
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