Finding $l^p$ norm of $2\times2$ matrix for $p\notin\{1,2,\infty\}$ Suppose I have the matrix $A=\begin{pmatrix}1 & 5 \\ 5 & 2\end{pmatrix}$ and I plot the image of the unit ball in $l^4$ under $A$.

How can I use this to determine the induced $4$-norm of $A$? I understand that $\|A\|_4=\underset{\|\vec{x}\|_4=1}{max}\|A\vec{x}\|_4$. So I think my answer is given by that vector on the transformed ball (i.e. lies on the red curve) which has the greatest $4$-norm. (Can someone confirm this is correct?)
Here is where I am confused. If I was working with the $l^2$ ball and $l^2$ norm, I could find the vector that achieves the induced $\|A\|_2$ norm. This is just the vector on the red curve whose Euclidean distance from the origin is greatest. The $\|A\|_2$ norm, then, is equal to this 'greatest distance'. But I cannot apply the same logic to the $l^4$ case.
My guess is that the vector $\vec{v}$ on the red curve at the greatest Euclidean distance from the origin is also the same vector that achieves the induced $\|A\|_4$ norm. If this were the case, then I could find the co-ordinates of this vector $\vec{v}$ and compute its $4$-norm by definition. But is this actually the case? It makes intuitive sense, but could someone explain why? If not, how can I go about finding this vector?
(Note: I am not trying to solve this analytically, I only need my answer correct to a few decimal places, hence my method) 

Consider a point $P$ on the red curve and $P'$ the intersection of the ray $OP$ with the blue curve. Then $\|OP'\|_4=1$ so $\|OP\|=\frac{OP}{OP'}$. Therefore, you need to find $P$ so that $\frac{OP}{OP'}$ is maximum.  Alrenatively, you need to find the smallest dilation of the blue curve that contains inside the red curve. Try some dilations ( it seems you use Geogebra) and see one that may just fit.
Added: Just played a bit with Geogebra and it seems that the curve $x^4 +y^4 = 6.57^4$ just barely contains the red curve. So the norm is approximately $6.57$. 

