Find the normal of slope 2/3 to the hyperbola 3x^2-4y^2=8.

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lfryerda | High School Teacher | (Level 2) Educator

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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`



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


which simplifies to 


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}`.


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