#### Question

Find the equations of all lines of slope zero and that are tangent to the curve \[y = \frac{1}{x^2 - 2x + 3}\] ?

#### Solution

Slope of the given tangent is 0.

\[\text { Let }\left( x_1 , y_1 \right)\text { be a point where the tangent is drawn to the curve} (1).\]

\[\text { Since, the point lies on the curve } . \]

\[\text { Hence }, y_1 = \frac{1}{{x_1}^2 - 2 x_1 + 3} . . . \left( 1 \right) \]

\[\text { Now,} y = \frac{1}{x^2 - 2x + 3}\]

\[ \Rightarrow \frac{dy}{dx} = \frac{\left( x^2 - 2x + 3 \right)\left( 0 \right) - \left( 2x - 2 \right)1}{\left( x^2 - 2x + 3 \right)^2} = \frac{- 2x + 2}{\left( x^2 - 2x + 3 \right)^2}\]

\[\text { Slope of tangent }=\frac{- 2 x_1 + 2}{\left( {x_1}^2 - 2 x_1 + 3 \right)^2}\]

\[\text { Given that }\]

\[\text { Slope oftangent= slope of the given line }\]

\[ \Rightarrow \frac{- 2 x_1 + 2}{\left( {x_1}^2 - 2 x_1 + 3 \right)^2} = 0\]

\[ \Rightarrow - 2 x_1 + 2 = 0\]

\[ \Rightarrow 2 x_1 = 2\]

\[ \Rightarrow x_1 = 1\]

\[\text { Now }, y = \frac{1}{1 - 2 + 3} = \frac{1}{2} \left[ \text { From }\left( 1 \right) \right]\]

\[ \therefore \left( x_1 , y_1 \right) = \left( 1, \frac{1}{2} \right)\]

\[\text { Equation oftangentis},\]

\[y - y_1 = m \left( x - x_1 \right)\]

\[ \Rightarrow y - \frac{1}{2} = 0 \left( x - 1 \right)\]

\[ \Rightarrow y = \frac{1}{2}\]