Tangent Planes and Surfaces (Calc 3)

Question

I am wondering if I am on the right track for the following question:

Find a for the plane $x+y+z=-1$ so that it is a tangent plane to the surface $z=x^2+ay^2$

I figured since you are given a tangent plane and a surface that you would need to parameterize the surface using the following equation: r(u,v)=x(u,v)i+y(u,v)j+z(u,v)k. You would next derive in terms of u and v so you would have the derivative of u(r(u)) and derivative of v(r(v)), which you would then use the cross product on r(u) X r(v).

Would this be able to work even though I am solving for the variable a?

Answer 1

Intuition: a tangent plane intersects a convex surface (e.g. a paraboloid) in exactly one point. If that's true, and substituting one equation into the other, we seek the only solution $(x,y)$ which satisfies $x^2+x+(ay^2+y+1)=0$. Applying the formula for the roots of a quadratic equation (twice) and demanding a single root, I got the condition for $a=\frac{1}{3}$.