Central bank loss function $$
L_t = \gamma(\pi_t - \pi_t^\otimes)^2 + \hat{Y}_t^2
$$

Central banks loss function is given by the equation above. This loss is increasing and convex in the distance from the inflation target, i.e. the marginal loss increases at a faster rate than the increase in the distance from the target.
In Figure 1, we can visualise the preferences using indifference curves, where the ellipses have the center being the desired point (0, $\pi_t^\otimes$ = Target inflation point.) (saturation point - bliss point)
It is hard for me to see how the different ellipses are indifference curves, or convex from in the difference from target?. Any explanation would help a lot.
The intuitive way


Draw the function $L_t$, the graph below is build with $\gamma = 0.8$ and $\pi_t^\otimes = 1$



Convex in this context means that the function $L_t$ looks like a bowl, just one well defined minimum. And if you cut it with a horizontal plane, the result is an ellipse.




  The formal way


Just for convenience, define the variable $p_t = \pi_t - \pi_t^\otimes$ so the function becomes

$$
L_t(p_t, Y_t) = \gamma p_t^2 + Y_t^2 \tag{B1}
$$

First note that if $\gamma &gt; 0$  then $L_t(p_t, Y_t) \ge 0$. To prove convexity you just need to show that for any pair $(p_{t,1},Y_{t,1})$ and $(p_{t,2},Y_{t,2})$ the condition

$$
L_t(t p_{t,1} + (1 - t)p_{t,2}, t Y_1 + (1 - t)Y_{t,2}) &lt; tL_t(p_{t,1}, Y_t) + (1 -t)L_t(p_{t,2}, Y_{t,2}) \tag{B2}
$$

holds for $ 0 &lt; t &lt; 1$. This is actually fairly easy to show

\begin{eqnarray}
L_t(t p_{t,1} + (1 - t)p_{t,2}, t Y_1 + (1 - t)Y_{t,2}) &amp;=&amp; \gamma[t p_{t,1} + (1 - t)p_{t,2}]^2 + (t Y_1 + (1 - t)Y_{t,2})^2 \\
&amp;\vdots &amp; \\
&amp;=&amp; t [\gamma p_{t,1}^2 + Y_{t,1}^2] + (1 - t) [\gamma p_{t,2}^2 + Y_{t,2}^2] \\
&amp;&amp;\quad + t(1 - t) [\gamma(p_{t,1} - p_{t,2})^2 + (Y_{t,1} - Y_{t,2})^2] \\
&amp;=&amp; tL_t(p_{t,1}, Y_t) + (1 -t)L_t(p_{t,2}, Y_{t,2}) \\
&amp;&amp;\quad \underbrace{t(1 - t)L_t(p_{t,1} - p _{t_2}, Y_{t,1} - Y_{t,2})}_{&gt; 0} \\
&amp;&lt; &amp; tL_t(p_{t,1}, Y_t) + (1 -t)L_t(p_{t,2}, Y_{t,2}) \tag{B3}
\end{eqnarray}

So $L_t$ is convex. To show that indifference curves are ellipses, just fix $L_t$ to a constant value, let's say $A &gt; 0$ then we have

$$
A = \gamma p_t^2+ Y_t^2 ~~\Rightarrow~~ \frac{p_t^2}{A/\gamma} + \frac{Y_t}{A} = 1 \tag{B4}
$$

which are just ellipses centered at $p_t = 0 = \pi_t - \pi_t^\otimes$ and $Y_t = 0$ with axis lengths $\sqrt{A/\gamma}$ and $\sqrt{A}$