Question

Find the equations of all lines having slope 0 which are tangent to the curve $y=\frac{1}{x^2-2x+3}$

Answer 1

The equation of the given curve is $y=\frac{1}{x^2-2x+3}$ ...(i)

The slope of the tangent to the given curve at any point (x,y) is give by

$\frac{dy}{dx}=\frac{-1}{(x^2-2x+3{)}^2}=0\Rightarrow \frac{d}{dx}(x^2-2x+3)=\frac{-(2x-2)}{(x^2-2x+3{)}^2}=\frac{-2(x-1)}{(x^2-2x+3{)}^2}$

For all tangents having slope 0, we must have $\frac{dy}{dx}=0$

$\Rightarrow \frac{-2(x-1)}{(x^2-2x+3{)}^2}=0\Rightarrow -2(x-1)=0\Rightarrow x=1$

When x=1, from Eq. (i), we get $y=\frac{1}{1^2-2\times 1+3}=\frac{1}{2}$

$\therefore$ The equation of tangent to the given curve at point $(1,\frac{1}{2})$ having slope = 0

$y-\frac{1}{2}=0(x-1)\Rightarrow y=\frac{1}{2}$

Hence, the equation of the required line is $y=\frac{1}{2}$.