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**deezy** Find pt on paraboloid $\displaystyle x = 5y^2 + 7z^2$, if it exists, at which the tangent plane is parallel to plane $\displaystyle -x + y + z = 3$.

Not completely sure how to approach this problem. I'm pretty sure it involves the gradient, so I set $\displaystyle f(x,y,z) = x - 5y^2 - 7z^2$ and found that gradient which was $\displaystyle \nabla f = i - 10yj - 14zk$.

Do I need the equation for the tangent plane of the paraboid? And are -1, 1, 1 the directional numbers for the plane $\displaystyle -x + y + z = 3$?