# What is the slope of the line normal to the tangent line of f(x) = x-sqrt(x^2+4)  at  x= 2 ?

May 14, 2017

Slope of normal is $\sqrt{2} x - y \left(1 - \sqrt{2}\right) - 6 \sqrt{2} + 6 = 0$

#### Explanation:

While the slope of tangent at $\left({x}_{1} , f \left({x}_{1}\right)\right)$ for $f \left(x\right)$ is given by value of first derivative at that point i.e. f((x_1),

as normal and tangent are perpendicular to each other (i.e. their product is $- 1$), slope of normal is $\frac{1}{f ' \left({x}_{1}\right)}$.

Now we have $f \left(x\right) = x - \sqrt{{x}^{2} + 4}$, and we are seeking tangent at $x = 2$ i.e. at $\left(2 , 2 - \sqrt{{2}^{2} + 4}\right)$ i.e. at $\left(2 , 2 - 2 \sqrt{2}\right)$.

Now $f ' \left(x\right) = 1 - \frac{2 x}{2 \sqrt{{x}^{2} + 4}} = 1 - \frac{x}{\sqrt{{x}^{2} + 4}}$

and slope of tangent is $f ' \left(2\right) = 1 - \frac{2}{2 \sqrt{2}} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}}$

and equation of tangent is $y - 2 + 2 \sqrt{2} = \frac{\sqrt{2} - 1}{\sqrt{2}} \left(x - 2\right)$

i.e. $\sqrt{2} y - 2 \sqrt{2} + 4 = \left(\sqrt{2} - 1\right) x - 2 \sqrt{2} + 2$

or $\left(\sqrt{2} - 1\right) x - \sqrt{2} y - 2 = 0$

and hence slope of normal is $\frac{\sqrt{2}}{1 - \sqrt{2}}$ and equation of normal is

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

or $y \left(1 - \sqrt{2}\right) - 2 {\left(1 - \sqrt{2}\right)}^{2} = \sqrt{2} x - 2 \sqrt{2}$

or $\sqrt{2} x - y \left(1 - \sqrt{2}\right) - 2 \sqrt{2} + 2 \left(3 - 2 \sqrt{2}\right) = 0$

or $\sqrt{2} x - y \left(1 - \sqrt{2}\right) - 6 \sqrt{2} + 6 = 0$

graph{((sqrt2-1)x-sqrt2y-2)(sqrt2x-y(1-sqrt2)-6sqrt2+6)(x-sqrt(x^2+4)-y)=0 [-5, 5, -2.5, 2.5]}