A) \[3x+27y=79\]
B) \[27\text{ }x-3y=79\]
C) \[27x+3y=79\]
D) \[3x-27y=79\]
Correct Answer: B
Solution :
Given curve is \[{{x}^{2}}y={{x}^{2}}-3x+6\] ?.(i) At \[x=3,\] \[{{3}^{2}}(y)={{3}^{2}}-3(3)+6\] \[\Rightarrow \] \[y=\frac{2}{3}\] On differentiating Eq. (i) w.r.t.x, we get \[2xy+{{x}^{2}}\frac{dy}{dx}=2x-3\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{2x-3-2xy}{{{x}^{2}}}\] \[\Rightarrow \] \[{{\left( \frac{dy}{dx} \right)}_{\left( 3,\frac{2}{3} \right)}}=\frac{6-3-2\times 3\times \frac{2}{3}}{{{3}^{2}}}=-\frac{1}{{{3}^{2}}}\] \[\therefore \] Equation of normal is \[y-\frac{2}{3}={{3}^{2}}(x-3)\] \[\Rightarrow \] \[3y-2=27(x-3)\] \[\Rightarrow \] \[27x-3y=79\]
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