# Thread:

1. ## 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?

2. 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?

3. 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:
$x\left(t \right)=x_{0}+f_{x}\left(x_{0},y_{0},z_{0} \right)$

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).
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