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. $27x-3y=79$
• 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
$y-\frac{2}{3}=3^2(x-3)$
$\Rightarrow$ $3y-2=27(x-3)$
$\Rightarrow$ $27x-3y=79$