Confidence band for simple linear regression - why the curve? I am really struggling to understand why confidence bands for regression lines have a curve to them. A few example plots showing curved CIs, taken from this post: Shape of confidence interval for predicted values in linear regression

I have read a bit and understand that the standard error for the sampling distribution of $\hat y$ at a given point ($\hat y^*$) is as follows:
$$ MSE \sqrt{\frac{1}{n} + \frac{(x^*-\bar x)^2}{\sum_{i=1}^n (x_i-\bar x)^2}} $$
Because the above equation has $(x^* - \bar x)^2$ in the numerator, the further we are from $\bar x$, the bigger the standard error will be, hence the curve being 'thinnest' when $x = \bar x$, and fatter as $x$ increases/decreases.
But why does that make sense?
My thinking is this: if one of the assumptions for a linear regression is homogeneity of variance for $Y$ across values of $X$, and the residuals from our regression are also equally distributed across values of $X$,  why would the variance of our sampling distributions of $\hat y$ not be the same at each value of $X$?
Clearly I am missing something!

Computing the sample variance of the estimate $\hat{y}$
The estimate for the mean $y$ (as a function of $x$) has the following function in terms of the predictions for coefficients $\alpha$ and $\beta$
$$\hat {y} = \hat{\alpha} + \hat{\beta} x$$
The standard error of $\hat{y}$ can be computed with the formula for the standard deviation or variance
$$Var(\hat\alpha + \hat\beta x) = Var(\hat\alpha) + x^2 Var(\hat\beta) - 2x  Cov(\hat\alpha,\hat\beta)$$
So this is a quadratic function that has a minimum at $x = \frac{Cov(\hat\alpha,\hat\beta)}{Var(\hat\beta)} = \bar{x_i}$ and this creates that funnel shape with the minimum at the mean of the datapoints $x_i$.
Intuitive
Let's try out several fits. We use the following data
$X_i$ is normal distributed. $Y_i$ is $0.8$ times $X_i$ with some added noise.
$$\begin{array}{}
X_i &\sim& N(0,1) \\ \epsilon_i &\sim& N(0,1) \\
Y_i &=& 0.8 X_i + 0.6 \epsilon_i
\end{array}$$
Result of $25$ simulations with each $15$ data points

When we combine all those different lines in a single plot then we get:

So here we might see intuitively why the confidence band becomes 'fatter' at the ends. The confidence is due to errors in the height of the line (parameter $\alpha$) and the slope of the line (parameter $\beta$). It is this latter one that makes the error larger towards the ends.

$\frac{Cov(\hat\alpha,\hat\beta)}{Var(\hat\beta)} = \bar{x} $ follows from the covariance matrix for the $\alpha$ and $\beta$ which is $\sigma (X^TX)^{-1}$. Which you could work out further by filling in all the terms...  But you could also argue that the minimum should be at $\bar{x}$ by transform the data matrix $X$ such that the the column vectors are perpendicular and the estimates $\hat\alpha$ and $\hat\beta$ have zero covariance.
