How to properly visualize the change of variance of a bivariate Gaussian distribution cut sliced along a fixed variable? I'm trying to understand the basics of Gaussian Distribution. I struggle to visualice how the variance of the conditional probability of say P (Y|X) changes when X is fixed (given X and Y have a joint gaussian distribution). So, I have two pictures , in picture A from many sources shows that the variances dont change but the mean does.  But it seems reasonable to me also to consider the conditional as a cut of de bivariate along one axis but now the variances does change, why is it so?  I'm I  wrongly thinking Picture B is the conditional?
Picture A

[Picture updated thanks to contribution]
Picture B
Cuts along one variable.  It seems that the variance (spread of black lines) increases or decreases according to a fixed X.


Suppose that $X$ and $Y$ are jointly normal with correlation coefficient $\rho \in (-1,1)$ and identical marginal $\mathcal N(0,1)$ distributions. Then, the joint density is
$$f_{X,Y}(x,y) = \frac{1}{2\pi \sqrt{1-\rho^2}}\exp\left[-\left.\left.\frac{1}{2(1-\rho^2)}\right(x^2-2\rho xy + y^2\right)\right]\tag{1}$$
and for any fixed value $x_0$ for $X$, we have that
$$f_{X,Y}(x_0,y) = \frac{1}{2\pi \sqrt{1-\rho^2}}\exp\left[-\left.\left.\frac{1}{2(1-\rho^2)}\right(x_0^2-2\rho x_0y + y^2\right)\right].\tag{2}$$
The OP has drawn pictures of Eq. $(2)$ for different values of $x_0$ and seems to think that these cross-sections of the joint pdf solid are the various conditional densities of $Y$ given that $X$ has those specific values. He seems to think that the variances of these conditional densities increase as $|x_0|$ increases because the cross-sections seem more spread out.  But, $f_{Y\mid X = x_0}(y\mid X=x_0)$, the conditional density of $Y$ given that $X=x_0$, is not given by $(2)$.  In fact, $f_{X,Y}(x_0,y)$ is a valid pdf only in rare circumstances. Note that, in general and without any assumptions about normality etc,
the integral $\int_{-\infty}^\infty f_{X,Y}(x_0,y) \,\mathrm dy$ equals $f_X(x_0)$ and so while $f_{X,Y}(x_0,y)$ is indeed nonnegative (as all pdfs must be), the area under the curve does not equal $1$ except when $x_0$ is such that $f_X(x_0)$ serendipitously happens to have value $1$. Note that $\dfrac{f_{X,Y}(x_0,y)}{f_X(x_0)}$, which we recognize immediately as the formula for the conditional pdf of $Y$ given that $X$ has value $x_0$, is indeed as valid pdf.  Thus,
\begin{align}
f_{Y\mid X = x_0}(y\mid X=x_0) &= \dfrac{f_{X,Y}(x_0,y)}{f_X(x_0)}\\
&= \dfrac{\frac{1}{2\pi \sqrt{1-\rho^2}}\exp\left[-\left.\left.\frac{1}{2(1-\rho^2)}\right(x_0^2-2\rho x_0y + y^2\right)\right]}{\frac{1}{ \sqrt{2\pi}}\exp\left[-\left.\left.\frac{1}{2}\right(x_0^2\right)\right]}\\
&= \frac{1}{\sqrt{1-\rho^2}\sqrt{2\pi}}\exp\left[-\left.\left.\frac{1}{2}\right(\frac{y-\rho x_0}{\sqrt{1-\rho^2}}\right)^2\right]
\end{align}
which shows that the conditional density of $Y$ given that $X=x_0$ is a normal density with fixed variance $1-\rho^2$ (not depending on the given value $x_0$ of $X$ at all) but with mean $\rho x_0$ which does vary with $x_0$.  This is what several of the comments on the OP's question are trying to point out:

the cross-section of the joint pdf solid at $x_0$ is not the conditional density of $Y$ given $X=x_0$ (though it is related). The conditional density has fixed variance (not depending on $x_0$) but variable mean (depending on $x_0$).

It is possible to read too much into the aphorism that the bivariate normal distribution is like a piece of bologna in the sense no matter how you slice it, it is still bologna. Every cross-section (not necessarily parallel to the axes) of the bivariate normal density solid is proportional to a normal density but is not exactly a normal density unless the cross-section passes through the mean point of the bivariate distribution.
