Question

The slope of normal to the curve $y=\sqrt{x^2-3x+3}$ at point, where its ordinate and abscissa are equal, is

Answer 1

Given curve : $y=\sqrt{x^2-3x+3}$
when ordinate will be equal to abscissa
i.e., $y=x$
$\Rightarrow x^2=x^2-3x+3$
$\Rightarrow x=1$
Now,
$\frac{dy}{dx}=\frac{2x-3}{2\sqrt{x^2-3x+3}}$
Slope of tangent at $x=1$
$\frac{dy}{dx}{|}_{x=1}=-\frac{1}{2}$
$\therefore$ slope of normal $=2$