# How do you find the equation of the tangent and normal line to the curve y=lnx at x=17?

Nov 2, 2016

Tangent Equation : $y = \frac{x}{17} + \ln 17 - 1$
Normal Equation : $y = - 17 x + \ln 17 + 289$

#### Explanation:

If $y = \ln x$ then $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{x}$

When $x = 17$
$\implies y = \ln 17$
$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{17}$

so the tangent passes through $\left(17 , \ln 17\right)$ and has gradient ${m}_{T} = \frac{1}{17}$

Using $y - {y}_{1} = m \left(x - {x}_{1}\right)$ the equation of the tangent is:

$y - \ln 17 = \frac{1}{17} \left(x - 17\right)$
$\therefore y - \ln 17 = \frac{x}{17} - 1$
$\therefore y = \frac{x}{17} + \ln 17 - 1$

The normal is perpendicular to the tangent, so the product of their gradients is -1 hence normal passes through $\left(17 , \ln 17\right)$ and has gradient ${m}_{N} = - 17$

so the equation of the normal is:
$y - \ln 17 = - 17 \left(x - 17\right)$
$\therefore y - \ln 17 = - 17 x + 289$
$\therefore y = - 17 x + \ln 17 + 289$