# Tangent planes for the function $f(x,y) = 1 - x^2 - y^2$

I have to solve the following statements, finding the tangent planes to $$f$$ and its points of tangency:

that contains the line in $$\mathbb{R}^3$$ that passes through the points $$(3,0,3)$$ and $$(0,-3,3)$$

and this other statement:

that contains the point $$(0,0,2)$$

For the first statement, I tried first making the parametric equation and I found that $$L(t) = (3-3t, -3t, 3)$$, and I also know that the equation of the plane is given by $$p(x,y) = \frac{\partial f}{\partial x}(x_0,y_0)(x-x_0) + \frac{\partial f}{\partial y}(x_0,y_0)(y-y_0) + f(x_0,y_0) = -2x_0 (x-x_0) -2y_0 (y-y_0) + (1 -x_0^2 - y_0^2)$$ but after that, I don´t see what else to do. Could you give me any hint for both statements?

• Do you know the formula of the tangent plane to a surface $z=f(x,y)$ at a generic point? Dec 12 '20 at 19:36
• I edited my post, I wrote the formula with partial derivatives. Dec 12 '20 at 19:38

The partial derivatives are:$$f'_x = -2x, \quad f'_y = -2y$$ Thus, we have \begin{align}z &= \frac{\partial f}{\partial x}(x_0,y_0)(x-x_0) + \frac{\partial f}{\partial y}(x_0,y_0)(y-y_0) + f(x_0,y_0) \\ &= -2x_0(x-x_0) - 2y_0(y-y_0) -x_0^2 - y_0^2 + 1\tag{1}\end{align} Since $$z(0,0) = 2$$, we have $$x_0^2 +y_0^2 + 1 = 2 \implies x^2_0 + y_0^2 = 1 \tag{2}$$ From this, $$(1)$$ simplifies to \begin{align}z&=-2x_0(x-x_0) - 2y_0(y-y_0) -x_0^2 - y_0^2 + 1 \\&= -2x_0x-2y_0y+x_0^2+y_0^2+1 \\&= -2x_0x-2y_0y + 2 \tag{3}\end{align}

For all points $$(x_0,y_0)$$ satisfying $$(2)$$, the equation $$(3)$$ gives a tangent plane that passes through $$(0,0,2)$$: For example, when $$x_0 = \sqrt{\frac{2}{3}}$$ and $$y_0 = \sqrt{\frac{1}{3}}$$ you get the following picture:

One more example with another random choice of a tangency point satisfying $$(2)$$:

• And how can you make this planes tangent to the function $f$? Dec 12 '20 at 20:16
• Can I say that the set of points where this plane is tangent is the intersection of $x^2 + y^2 = 1$ with the graph of $f$? Dec 12 '20 at 20:22
• @SocietyViper See the last addition Dec 12 '20 at 20:26
• Okay, that's what I was looking at. And for the first statement, do you have any idea? I've been trying for a while, but can't find a way to get to the result. Dec 12 '20 at 20:27
• @SocietyViper For the first part, the above answer and this should help: math.stackexchange.com/questions/1401716/… Dec 12 '20 at 20:28

Given

$$\cases{ f(x,y,z)=x^2+y^2+z-1=0\\ \pi(x,y,z) = a x+b y + z c + d=0\\ p=(x,y,z)\\ p_1=(3,0,3)\\ p_2=(0,-3, 3) }$$

Construct $$\pi(x,y,z)$$ such that it contains the line $$L \to p = p_1+\mu(p_2-p_1)$$ and is tangent to $$f(x,y,z)$$.

At the tangency point $$p_0$$ we have

$$\cases{ \nabla f(p_0) = \lambda \nabla \pi(x_0)\\ f(p_0) = 0\\ \pi(p_0) = 0\\ \pi(p_1) = 0\\ (p_2-p_1)\cdot\nabla \pi(p_0) = 0\\ \|\nabla \pi(p_0)\|=1 }$$

Here

$$\cases{ \nabla f(p_0) = (2x_0,2y_0,1)\\ \nabla\pi(p_0) = (a,b,c) }$$

so we have $$8$$ equations on the $$8$$ unknowns $$x_0,y_0,z_0,a,b,c,d,\lambda$$. Solving those equations we have two solutions:

$$p_0 = \left\{ \begin{array}{ccc} (\frac{1}{2} \left(3-\sqrt{13}\right), & \frac{1}{2} \left(\sqrt{13}-3\right), & 3 \sqrt{13}-10) \\ ( \frac{1}{2} \left(3+\sqrt{13}\right), & -\frac{1}{2} \left(3+\sqrt{13}\right), & -10-3 \sqrt{13})\\ \end{array} \right.$$

and the planes are characterized by

$$(a,b,c,d) = \left\{ \begin{array}{cccc} (-\sqrt{\frac{2}{51} \left(9-\sqrt{13}\right)}, & \sqrt{\frac{2}{51} \left(9-\sqrt{13}\right)}, & \frac{1}{68} \left(\sqrt{102} \left(9-\sqrt{13}\right)^{3/2}-8 \sqrt{102 \left(9-\sqrt{13}\right)}\right), & \frac{1}{204} \left(12 \sqrt{102 \left(9-\sqrt{13}\right)}-\sqrt{102} \left(9-\sqrt{13}\right)^{3/2}\right)) \\ (-\sqrt{\frac{2}{51} \left(9+\sqrt{13}\right)}, & \sqrt{\frac{2}{51} \left(9+\sqrt{13}\right)} ,& \frac{1}{68} \left(\sqrt{102} \left(9+\sqrt{13}\right)^{3/2}-8 \sqrt{102 \left(9+\sqrt{13}\right)}\right), & \frac{1}{204} \left(12 \sqrt{102 \left(9+\sqrt{13}\right)}-\sqrt{102} \left(9+\sqrt{13}\right)^{3/2}\right)) \\ \end{array} \right.$$

Attached a plot showing in red the given line and in black the tangentcy points.