Sketch the unit ball $B(0, 1)$ in $\mathbb{R}^2$ equipped with the norm $||(x,y)||=\sqrt{2x^2+3y^2}$

Question

Sketch the unit ball $B(0, 1)$ in $\mathbb{R}^2$ that has the norm:

$||(x,y)||=\sqrt{2x^2+3y^2}$

Im ok at sketching unit balls normally but the norm given has stumped me in making progress with this question so any help will be appreciated.

Answer 1

The boundary of that unit ball would be the points $x,y$ so that $$\sqrt{2x^2+3y^2}=1 \iff 2x^2+3y^2=1$$

As they've already pointed out in the comments, that's the equation of an ellipse. It's easier to sketch it if you give some easy values to $x$ and $y$. For example, $x=0 \Rightarrow y=\frac{\pm1}{\sqrt3}$, $y=0 \Rightarrow x=\frac{\pm1}{\sqrt2}$...

Answer 2

It's an ellipse passing through $||v_i||$ for $v_i\in\big\{(1,0),(0,1),(-1,0),(0,-1)\big\}$.

Note: For generating such pictures by yourself you can use Geogebra and enter 1 = sqrt(2x^2 + 3y^2) in the bar at the bottom of the window.