# How do you find the equations for the normal line to x^2/32+y^2/8=1 through (4,2)?

Dec 23, 2016

The normal line to the curve through the point $\left(4 , 2\right)$ has equation:

$y = 2 x - 6$

#### Explanation:

For a curve in implicit form:

$F \left(x , y\right) = 0$

the normal line to the point $\left(\overline{x} , \overline{y}\right)$ is given by the formula:

$\frac{\partial F}{\partial y} \left(\overline{x} , \overline{y}\right) \left(x - \overline{x}\right) - \frac{\partial F}{\partial x} \left(\overline{x} , \overline{y}\right) \left(y - \overline{y}\right) = 0$

We have:

$F \left(x , y\right) = {x}^{2} / 32 + {y}^{2} / 8 - 1$

$\frac{\partial F}{\partial x} = \frac{x}{16}$

$\frac{\partial F}{\partial y} = \frac{y}{4}$

so that the line normal to the curve in $\left(4 , 2\right)$ has equation:

$\frac{2}{4} \left(x - 4\right) - \frac{4}{16} \left(y - 2\right) = 0$

or multypling everything by 4 to have integral coefficients:

$2 \left(x - 4\right) - \left(y - 2\right) = 0$

$2 x - 8 - y + 2 = 0$

$y = 2 x - 6$ Dec 24, 2016

$y = 2 x - 6$

#### Explanation:

The tangent at $\left(x ' , y '\right)$ to a conic section is given by a substitution rule, which in this case yields $\frac{x x '}{32} + \frac{y y '}{8} = 1$, which for $x ' = 4$, $y ' = 2$ may be re-written as $x + 2 y = 8$. Consequently the normal will be $2 x - y = c$ (using ${m}_{1} \times {m}_{2} = - 1$) where $c$ is evaluated by substituting $x = 4$, $y = 2$, giving $8 - 2 = 6$. Hence the normal is $2 x - y = 6$, re-arrangeable to $y = 2 x - 6$.

This works for any conic section anywhere, with suitable further substitution rules. More complex rules give chords.