Consider the surfaces $$x^2+y^2-2z^2=2$$ and $$(x^2+y^2)z=4$$
Prove that, for any point $p_0$, on the intersection of these two surfaces, the two tangent planes at $p_0$ are orthogonal.
Would it be satisfactory to find where the surfaces intersect and then find their respective gradient vectors? I know the tangent planes will be orthogonal if the normal line from one plane is on the tangent plane of the other which occurs if: $$\nabla f(x,y,1)\cdot \nabla g(x,y,1)=0.$$
Any help would be appreciated.
Best Answer
first the intersection points are: $$r^{2}=2+2z^{2}$$ $$r^{2}=4/z$$ $$2+2z^{2}=4/z$$ $$z=1$$ for $z=1$ we have: $$ r=2$$ so the intersection point is: $$x=2cos\theta ,y=2sin\theta, z=1$$
The normal of the tangent plane of the first surface: $$ n=(2x,2y,-4z)$$
the normal of the tangent plane of the second one is:
$$n=(2xz,2yz,x^{2}+y^{2})$$
the inner product of these normal vectors are: $$<n_{1}.n_{2}>=(4x^{2}z+4y^{2}z-4z(x^{2}+y^{2})$$
so in z=1 we have:$$ <n_{1}.n_{2}>=0$$ Notice that the inner product is always zero. I just wanted to show that these surfaces have intersection.