Find the slope of the relation `3x^2-4y^2=8` using implicit differentiation, which gives `6x-4y cdot y'=0`. Now we can solve for `y'`, which gives

`4y cdot y'=6x`

`y'={6x}/{4y}`

`y'={3x}/{2y}`

which means when the slope is `2/3`, then the point is when

`2/3={3x}/{2y}`

which simplifies to

`x={4y}/9`

We can substitute this into the original relation to give `16y^2-4y^2=8`.

This simplifies to `12y^2=8`, which has the solutions `y=+-1/sqrt 3`.

The two points with this solution are `(4/{9sqrt 3}, 1/sqrt 3)` and `(-4/{9sqrt 3}, -1/sqrt 3)`.

The normal line has a slope of `-3/2` which is the negative reciprocal of `2/3`.

Therefore the two normal lines are found by substituting each point and the normal slope into `y=mx+b`.

**This gives the normal lines `y=-3/2 x+5/{3sqrt 3}` and `y=-3/2x-5/{3sqrt3}`.**