Thread: Tangent and normal

to the curve, x^2 - sqrt(3) xy + 2y^2 = 5

I got the slope to equal 0.

Can someone help me out please?

Originally Posted by Morgan82

to the curve, x^2 - sqrt(3) xy + 2y^2 = 5

I got the slope to equal 0.

Can someone help me out please?

What exactly are you trying to do?

The equation for the tangent plane at a point is:


Find your partial derivatives, plug everything in, and simplify.

To find the normal line to this tangent plane, you then want to define x, y, and z parametrically. They will all have the form:


except with the appropriate variable in place of x. You then solve each equation for t and set all 3 equations equal (since t=t=t, this must be true).