Question

The equation of the normal to the curve $x^3+y^3=6xy$ at the point $(3,3).$

• A. $x+y-6=0$
• B. $-x-y+6=0$
• C. $x-y=0$
• D. $x+y=0$

Answer 1

The correct option is C $x-y=0$

$\frac{dy}{dx}=-\frac{fx}{fy}=-\frac{3x^2-6y}{3y^2-6x}=-1$ at $(3,3)$
$\therefore$ Required normal is $y-3=1(x-3)$ or $x-y=0$