# Tangent plane to a surface parallel to another plane

Find all the points $$(x_0,y_0,z_0)$$ on the surface with equation $$z=x^2+y^2$$

at which the tangent plane is parallel to the plane $$x+y=7$$ After finding the answer in this part, give a geometric interpretation.

I know the general equation of a tangent plane to a surface is $$z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$$

But how do I make the coeffecient of $$z$$ be $$0$$ ? And I don't think there exists any point on the surface such that the tangent plane is parallel to $$x+y=7$$ since this plane is perpendicular to the $$xy$$-plane, and cuts the cone shape $$z=x^2+y^2$$. But how do I write it out? Thank you guys!

• I suggest working this exercise backwards: start from the geometric interpretation. What’s interesting about the plane $x+y=7$? What sort of a surface is $z=x^2+y^2$?
– amd
Commented May 25, 2020 at 18:14
• The former is just a 2D-plane perpendicular to $xy$-plane, the latter is a cone shaped thing . Commented May 26, 2020 at 9:12
• So is there any point at which the side of the “cone-shaped thing” is vertical?
– amd
Commented May 26, 2020 at 22:28

Given a surface in the form $$f(x,y,z)=0$$, the normal vector to this surface at the point $$(x,y,z)$$ is simply given by $$\nabla f$$.

So the normal to $$z=x^2+y^2$$, or equivalently $$x^2+y^2-z=0$$, is given by $$(2x,2y,-1)$$.

The normal to the plane $$x+y=7$$ is $$(1,1,0)$$.

Thus, these two normals can never be parallel, hence there does not exist a point where their tangent planes are parallel.