# Find the equation of all lines having slope $$0$$ that are tangents to the curve $y=\large \frac{1}{x^2-2x+3}$

$\begin{array}{1 1} y = 0 \\ y=\frac{-1}{2} \\y=\frac{1}{2} \\ \text{No tangent} \end{array}$

Toolbox:
• If $y=f(x)$,then $\big(\large\frac{dy}{dx}\big)_P$=slope of the tangent to $y=f(x)$ at the point $P$.
• Equation of line with given two points is $\large\frac{x-x_1}{x_2-x_1}=\large\frac{y-y_1}{y_2-y_1}$
Step 1:
Given : $y=\large\frac{1}{x^2-2x+3}$
Differentiating w.r.t $x$ we get,
$\large\frac{dy}{dx}=\frac{(x^2-2x+3).0-1(2x-2)}{(x^2-2x+3)^2}$
$\large\frac{dy}{dx}=\frac{-(2x-2)}{(x^2-2x+3)^2}$
Step 2:
But the slope is given as 0.
Therefore $\large\frac{-(2x-2)}{(x^2-2x+3)^2}$$=0 \Rightarrow -2(x-1)=0 \Rightarrow x=1 Step 3: When x=1,y=\large\frac{1}{1^2-2(1)+3} \Rightarrow \large\frac{1}{2} Therefore the tangent at (1,\large\frac{1}{2}) having slope 0 is (y-y_1)=m(x-x_1) \Rightarrow (y-\large\frac{1}{2})=$$0(x-1)$