# Find the equations of all lines having slope 0 which are tangent to the curve .

Question:

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

Solution:

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

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

$\frac{d y}{d x}=\frac{-(2 x-2)}{\left(x^{2}-2 x+3\right)^{2}}=\frac{-2(x-1)}{\left(x^{2}-2 x+3\right)^{2}}$

If the slope of the tangent is 0, then we have:

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

$\Rightarrow-2(x-1)=0$

$\Rightarrow x=1$

When $x=1, y=\frac{1}{1-2+3}=\frac{1}{2}$

$\therefore$ The equation of the tangent through $\left(1, \frac{1}{2}\right)$ is given by,

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

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

$\Rightarrow y=\frac{1}{2}$

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