# How to find the analytical expression for the supporting hyperplane?

In Stephen Boyd & Lieven Vandenberghe's Convex Optimization book we have following set

$$\{x\in \mathbb R^2_+ \mid x_1x_2\geq1\}$$

Now the analytical expression for the supporting hyperplane is given as follows $$\frac{x_1}{t^2}+x_2=\frac{2}{t}.$$ I know that the supporting hyperplane at the boundary point $x_0$ is $a^Tx=a^Tx_0$. (The boundary of the above set is $(t,1/t)$). Further, I know that at $x_0$ the supporting hyperplane is trangent to the set. The tangent to the set at any boundary point is $-\frac{1}{t^2}$. I want to know how we obtain the following analytical expression for the supporting hyperplane $$\frac{x_1}{t^2}+x_2=\frac{2}{t}$$ Any help in this regard will be much appreciated. Thanks in advance.

Consider the set $C$ to be the epigraph of $f$: $$C = \{(x,r) \mid f(x)\leq r\}.$$ An outer normal vector at $(t,f(t))$ is given by $$\big(f'(t),-1\big)$$ and the supporting hyperplane is described by $$0=\langle (f'(t),-1),(x,f(x))-(t,f(t))\rangle.$$ Now if your case, $f(x)=1/x$ and so $f'(x)=-1/x^2$.
Consider the point $(t,1/t)$ fixed, and in the boundary of $C$ for your choice. Then $(x_1,x_2)=(x,f(x))$ is in the graph, where $x=x_1$ and $x_2=1/x_1$. In this case, \begin{align*}0&=\langle (f'(t),-1),(x,f(x))-(t,f(t))\rangle\\ &= \langle (-1/t^2,-1),(x_1,x_2)-(t,1/t)\rangle\\ &=(-1/t^2)(x_1-t)+(-1)(x_2-1/t),\\ &=\frac{-x_1}{t^2}+\frac{1}{t}-x_2+\frac{1}{t}, \end{align*} which turns in what you have.
• Thank you so much for your answer. I think your answer is a bit difficult to understand therefore I try to explain my own reasoning below. Pick a point on the boundary $(t,1/t)$. The tangent to the boundary is $(1,-1/t^2)$. The normal vector associated with the hyperplane must be normal to the tangent vector. Let the normal vector to be $a^T=[a_1 a_2]$ then we must have $<a^T,(1,-1/t^2)>=0$ which results in $a_1=a_2/t^2$. Now if we pick $a_2=1$ then we know that product of $a^T$ with the reference point $(t,1/t)$ is equal to the product of $a^T$ with $(x_1,x_2)$. Therefore ... – Frank Moses May 15 '18 at 6:02
• $$a^T[x_1 x_2]^T=a^T[t 1/t]^T$$ $$x_1/t^2+x_2=2/t$$. Is this explanation wrong? – Frank Moses May 15 '18 at 6:08