show that the curves 3x^2+2x-3y^2=1 and 6xy+2y=0 are orthogonal (i.e. there slopes everywhere are perpendicular)

thank you, your help is very appreciated!

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- Dec 15th 2008, 02:51 PMmxhockey140curves
show that the curves 3x^2+2x-3y^2=1 and 6xy+2y=0 are orthogonal (i.e. there slopes everywhere are perpendicular)

thank you, your help is very appreciated! - Dec 15th 2008, 03:24 PMbilla
If you take the derivatives of both curves wrt x then the derivative of the first curve is (3x+1)/(3y) and the derivative of the second curve wrt x is

(-3y)/(3x+1). So the slopes are opposite reciprocals. This means they are perpendicular. Hope this helps, someone else will hopefully explain it better. - Dec 15th 2008, 03:44 PMmxhockey140
im sorry i had a typo. the first equation i forgot to put = 1, i dont know if that changes anything at all or not..

- Dec 15th 2008, 03:57 PMbilla
nope because the derivative of a constant is still zero

- Dec 15th 2008, 04:29 PMmxhockey140
thank you!