Question

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

Answer 1

The normal line to the curve through the point `(4,2)` has equation:

`y=2x-6`

Explanation:

For a curve in implicit form:

`F(x,y) = 0`

the normal line to the point `(barx,bary)` is given by the formula:

`(delF)/(dely)(barx,bary)(x-barx)-(delF)/(delx)(barx,bary)(y-bary) = 0`

We have:

`F(x,y) = x^2/32+y^2/8-1`

`(delF)/(delx) = x/16`

`(delF)/(dely) = y/4`

so that the line normal to the curve in `(4,2)` has equation:

`2/4(x-4)-4/16(y-2)=0`

or multypling everything by 4 to have integral coefficients:

`2(x-4)-(y-2) = 0`

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

`y=2x-6`

Answer 2

`y=2x-6`

Explanation:

The tangent at `(x',y')` to a conic section is given by a substitution rule, which in this case yields `(x x')/32+(y y')/8=1`, which for `x'=4`, `y'=2` may be re-written as `x+2y=8`. Consequently the normal will be `2x-y=c` (using `m_1 xx m_2=-1`) where `c` is evaluated by substituting `x=4`, `y=2`, giving `8-2=6`. Hence the normal is `2x-y=6`, re-arrangeable to `y=2x-6`.

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