# How do you find the equations for the normal line to y=x^2 through (2,4)?

Dec 13, 2016

$y = - \frac{1}{4} x + \frac{9}{2}$

#### Explanation:

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

so If $y = {x}^{2}$ then differentiating wrt $x$ gives us:

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

When $x = 2 \implies y = {2}^{2} = 4$ (so $\left(2 , 4\right)$ lies on the curve)
and $\frac{\mathrm{dy}}{\mathrm{dx}} = \left(2\right) 2 = 4$

So the normal we seek passes through $\left(2 , 4\right)$ ad has gradient $- \frac{1}{4}$ so using $y - {y}_{1} = m \left(x - {x}_{1}\right)$ the equation we seek is;

$y - 4 = - \frac{1}{4} \left(x - 2\right)$
$\therefore y - 4 = - \frac{1}{4} x + \frac{1}{2}$
$\therefore y = - \frac{1}{4} x + \frac{9}{2}$

We can confirm this graphically: