A) \[{{x}^{2}}=2y+1-{{e}^{2y}}\]
B) \[{{x}^{2}}=2y-1+{{e}^{2y}}\]
C) \[{{x}^{2}}=2y+1-{{y}^{-2y}}\]
D) \[{{x}^{2}}=-2y+1-{{e}^{2}}y\]
Correct Answer: A
Solution :
\[y-{{y}_{1}}=-\frac{1}{\left( \frac{dy}{dx} \right)}(x-{{x}_{1}})\] Normal at \[P({{x}_{1}},\,{{y}_{1}})\]. \[Q\left( 0,\,{{y}_{1}}+{{x}_{1}}\frac{dx}{dy} \right)\] Midpoint of PQ is \[\left( \frac{{{x}_{1}}}{2},\,{{y}_{1}}+\frac{{{x}_{1}}}{2}\frac{dx}{dy} \right)\] which lies on \[{{x}^{2}}=\frac{y}{2}\]. \[\Rightarrow \] \[\frac{x_{1}^{2}}{4}=\frac{{{y}_{1}}}{2}+\frac{{{x}_{1}}}{4}\frac{dx}{dy}\] \[\Rightarrow \] \[x\frac{dx}{dy}-{{x}^{2}}=-2y\] \[{{x}^{2}}=t\Rightarrow \,2x\frac{dx}{dy}=\frac{dt}{dy}\] \[\Rightarrow \] \[\frac{dt}{dy}-2t=-4y\] \[\Rightarrow \] \[I.F.={{e}^{-2y}}\] \[\Rightarrow \] \[t\cdot \,{{e}^{-2y}}=-4\int_{{}}^{{}}{y\cdot \,{{e}^{-2y}}dy+c}\] \[\Rightarrow \] \[{{x}^{2}}{{e}^{-2y}}=-4\left( y\cdot \,\frac{{{e}^{-2y}}}{-2}-\frac{1}{2}\cdot \,\frac{{{e}^{-2y}}}{2} \right)+c\] \[\Rightarrow \] \[{{x}^{2}}=2y+1-{{e}^{2y}}\]
You need to login to perform this action.
You will be redirected in
3 sec
