# All tangent planes pass through a fixed point, what can I get?

I am faced with this problem:

Problem $$\Gamma: z=f(x,y)$$ is a surface in $$\mathbb{R}^3$$ and $$f(x,y)$$ has continuous derivatives on $$\mathbb{R}^2$$. All tangent planes of $$\Gamma$$ pass through a fixed point $$(a,b,c)$$. Prove that $$f(t(x-a)+a,t(y-b)+b)-c=t(f(x,y)-c), \forall t>0.$$ (Hint: What is the solution to $$\frac{y'(t)}{y(t)}=\frac{1}{t}?$$)

My Solution(unfinished) I have transform the aim formula into $$f(tx+(1-t)a,ty+(1-t)b)=tz+(1-t)c, \forall t>0,$$ which implies a line going through a random point $$P\in \Gamma$$ and the fixed point$$(a,b,c)$$ totally lies in $$\Gamma$$. Without losing generality, We take $$(a,b,c)=(0,0,0)$$ and the aim formula turns into $$f(tx,ty)=tf(x,y), \forall t>0.$$

The formula of tangent planes at a specific point $$(x_0,y_0,z_0)\in \Gamma$$ is $$\frac{\partial f\left( x_0,y_0 \right)}{\partial x}\left( x-x_0 \right) +\frac{\partial f\left( x_0,y_0 \right)}{\partial y}\left( y-y_0 \right) -\left( z-z_0 \right) =0,$$ As $$(0,0,0)$$ lies in it, we have $$\frac{\partial f\left( x_0,y_0 \right)}{\partial x}\left( 0-x_0 \right) +\frac{\partial f\left( x_0,y_0 \right)}{\partial y}\left( 0-y_0 \right) -\left( 0-z_0 \right) =0,$$ $$\Rightarrow x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=z=f\left( x,y \right) , \forall \left( x,y \right) \in \mathbb{R}^2.$$ This hints me at Euler Formula of homogeneous functions, but what we need to do is to prove the converse. We have $$\frac{\mathrm{d}f\left( tx,ty \right)}{\mathrm{d}t}=\frac{\partial f}{\partial tx}\frac{\partial tx}{\partial t}+\frac{\partial f}{\partial ty}\frac{\partial ty}{\partial t}=x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=f\left( x,y \right) ,$$ Integrating on both sides of the equation, we have $$f(tx,ty)=tf(x,y)+C_0$$ where $$C_0$$ is an arbitrary constant in $$\mathbb{R}$$.

Here comes the tricky part: how to eliminate the constant $$C_0$$, i.e., to determine $$C_0=0$$?

I'm not sure I understand all your thoughts, but as I see it you already have all relevant formulas, just not in a straightforward order. Or perhaps completing your proof is as easy as inserting $$t=1$$ in your last formula, $$f(x,y)=f(x,y)+C_0$$.

The idea is that you restrict $$f$$ to the line from $$(a,b)$$ through some fixed $$(x,y)$$. $$g(t)=f(a+t(x-a),b+t(y-b)).$$ The visual picture is that as a scalar function $$g$$ inherits the property that all of its tangents meet in $$(0,c)$$. This is only possible if $$g$$ itself is a linear function with $$g(0)=c$$. Then $$g(t)=g(0)+t(g(1)-g(0))$$ which is the claimed formula.

In detail, the time derivative of $$g$$ is $$g'(t)=∂_xf(a+t(x-a),b+t(y-b))(x-a)+∂_yf(a+t(x-a),b+t(y-b))(y-b)$$ By the tangent plane condition you have $$f(x(t),y(t))+∂_xf(x(t),y(t))(a-x(t))+∂_yf(x(t),y(t))(b-y(t))=c$$ Inserting $$x(t)=a+t(x-a)$$, $$y(t)=b+t(y-b)$$ results in the more compact formula $$g(t)-tg'(t)=c.$$ This has an integrating factor $$-t^{-2}$$, so that $$\frac{g(1)}{1}-\frac{g(t)}{t}=\frac{c}1-\frac{c}{t} \\ \implies f\bigl(a+t(x-a),b+t(y-b)\bigr)-c=t\,(f(x,y)-c)$$

• You are right haha... t=1 really works! How did I miss that-_-#! And thaaaanks for your detailed answer! It actually makes the problem clearer. :) Oct 8 '20 at 9:46

Hint.

From the tangency condition

$$\frac{\partial f\left( x,y \right)}{\partial x}\left( x-a \right) +\frac{\partial f\left( x,y \right)}{\partial y}\left( y-b \right) -\left( f(x,y)-c \right) =0$$

solving this PDE we get at

$$f(x,y) = (x-a)\Phi\left(\frac{y-b}{x-a}\right)+c$$

now we can state that it is true

$$f(a+t(x-a),b+t(y-b))-c - t(f(x,y)-c) = 0$$

• Thank you for your hint! Could you explain a bit about $\phi$ function, please? Is it a function in PDE course? I haven't learnt about PDE so far... Oct 8 '20 at 14:06
• Well. It is an alternative approach to the question. The advantage is that you can get the functional structure to the ruled surface $f(x,y)$. Wait until you have enough knowledge in PDE's to enjoy the beauty of this result. Oct 8 '20 at 16:54
• I will keep it in mind. Many thanks! Oct 9 '20 at 6:31