# How do you find the equation of the tangent and normal line to the curve y=1/x at x=2?

Oct 6, 2016

Step 1 - Find the value of $y$ when $x = 2$:
$x = 2 \implies y = \frac{1}{2}$

Step 2 - Find the gradient of the tangent at $x = 2$:

$y = \frac{1}{x}$, $\therefore y = {x}^{-} 1$
So, $\frac{\mathrm{dy}}{\mathrm{dx}} = \left(- 1\right) {x}^{-} 2 = - \frac{1}{x} ^ 2$

When $x = 2 \implies \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2} ^ 2 = - \frac{1}{4}$

Step 3 - Eq'n of the Tangent when $x = 2$:
The tangent passes through $\left(2 , \frac{1}{2}\right)$ and has gradient $- \frac{1}{4}$
Using $y - {y}_{1} = m \left(x - {x}_{1}\right)$ we have the equation of the tangent is given by:
$y - \frac{1}{2} = - \frac{1}{4} \left(x - 2\right)$
$\therefore 4 y - 2 = - 1 \left(x - 2\right)$ (multiplying by 4)
$\therefore 4 y - 2 = 2 - x$
$\therefore 4 y + x = 4$

Step 4 - Eq'n of the Normal when $x = 2$:
Normal is perpendicular to Tangent, so
gradient of Normal = $- \frac{1}{- \frac{1}{4}} = 4$

The normal passes through $\left(2 , \frac{1}{2}\right)$ and has gradient $4$
Using $y - {y}_{1} = m \left(x - {x}_{1}\right)$ we have the equation of the normal is given by:
$y - \frac{1}{2} = 4 \left(x - 2\right)$
$\therefore 2 y - 1 = 8 \left(x - 2\right)$ (multiplying by 2)
$\therefore 2 y - 1 = 8 x - 16$
$\therefore 2 y - 8 x = - 15$