Question

The equation of normal to the curve $ {{x}^{2}}y={{x}^{2}}-3x+6 $ at the point with abscissa $ x=3 $ is

• A. $ 3x+27y = 75 $
• B. $ 27\text{ }x-3y=79 $
• C. $ 27x+3y = 75 $
• D. $ 3x-27y = 75 $

Answer 1

Answer: (B)

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 $