# How do you find the equation of the tangent and normal line to the curve y=6-x^2 at (2,2)?

Mar 22, 2017

Tangent:

$y = - 4 x + 10$

Normal:

$y = \frac{1}{4} x + \frac{3}{2}$

#### Explanation:

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. The normal is perpendicular to the tangent so the product of their gradients is $- 1$

We have:

$y = 6 - {x}^{2}$

First let us check that $\left(2 , 2\right)$ lies on the curve:

$x = 2 \implies y = 6 - 4 = 2$

Then differentiating wrt $x$ gives us:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x$

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

So the tangent passes through $\left(2 , 2\right)$ and has gradient ${m}_{T} = - 4$, and the normal has gradient ${m}_{N} = \frac{1}{4}$ so using the point/slope form $y - {y}_{1} = m \left(x - {x}_{1}\right)$ the equation we seek are;

Tangent:

$y - 2 = - 4 \left(x - 2\right)$
$\therefore y - 2 = - 4 x + 8$
$\therefore y = - 4 x + 10$

Normal:

$y - 2 = \frac{1}{4} \left(x - 2\right)$
$\therefore y - 2 = \frac{1}{4} x - \frac{1}{2}$
$\therefore y = \frac{1}{4} x + \frac{3}{2}$

We can confirm this solution is correct graphically: