What is the equation of the line that is normal to f(x)= sqry(x-2)  at  x=5 ?

May 23, 2017

Equation of tangent is $x - 2 \sqrt{3} y + 1 = 0$

Explanation:

We are seeking equation at $x = 5$ on the curve $f \left(x\right) = \sqrt{x - 2}$ i.e. at $\left(5 , \sqrt{3}\right)$ as $f \left(5\right) = \sqrt{5 - 2} = \sqrt{3}$

As slope of tangent at $f \left(5\right)$ is $f ' \left(5\right)$ let us first find $f ' \left(x\right)$

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

$\frac{1}{2 \sqrt{5 - 2}} = \frac{1}{2 \sqrt{3}}$

and equation of tangent is $y - \sqrt{3} = \frac{1}{2 \sqrt{3}} \left(x - 5\right)$

or $x - 2 \sqrt{3} y + 1 = 0$

graph{(x-2sqrt3y+1)(y-sqrt(x-2))=0 [0.81, 10.81, -0.56, 4.44]}