Question

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

Answer 1

Slope of the given tangent is 0.

$$\begin{gathered} \text{Let}\left(x_1,y_1\right)\text{be a point where the tangent is drawn to the curve (1).} \\ \mathrm{Since},\mathrm{the}\mathrm{point}\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{curve}. \\ \mathrm{Hence},y_1=\frac{1}{{x_1}^2-2x_1+3}...\left(1\right) \\ \mathrm{Now},y=\frac{1}{x^2-2x+3} \\ \Rightarrow \frac{dy}{dx}=\frac{\left(x^2-2x+3\right)\left(0\right)-\left(2x-2\right)1}{{\left(x^2-2x+3\right)}^2}=\frac{-2x+2}{{\left(x^2-2x+3\right)}^2} \\ \text{Slope of tangent=}\frac{-2x_1+2}{{\left({x_1}^2-2x_1+3\right)}^2} \\ \text{Given that} \\ \text{Slope of}\mathrm{tangent}\text{= slope of the given line} \\ \Rightarrow \frac{-2x_1+2}{{\left({x_1}^2-2x_1+3\right)}^2}=0 \\ \Rightarrow -2x_1+2=0 \\ \Rightarrow 2x_1=2 \\ \Rightarrow x_1=1 \\ \mathrm{Now},y=\frac{1}{1-2+3}=\frac{1}{2}\left[\mathrm{From}\left(1\right)\right] \\ \therefore \left(x_1,y_1\right)=\left(1,\frac{1}{2}\right) \\ \text{Equation of}\mathrm{tangent}\text{is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-\frac{1}{2}=0\left(x-1\right) \\ \Rightarrow y=\frac{1}{2} \end{gathered}$$