13. Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ passes through the point $P\left(\frac{\sqrt{6}}{2}, \frac{1}{2}\right)$, and the eccentricity is $\frac{\sqrt{2}}{2}$, with a moving point $M(2, t)(t>0)$.
(1) Find the standard equation of the ellipse;
(2) Find the equation of the circle with $O M$ as its diameter and which is intersected by the line $3 x-4 y-5=0$ to form a chord of length 2;
(3) Let $F$ be the right focus of the ellipse. Draw a perpendicular from point $F$ to $O M$ which intersects the circle with $O M$ as its diameter at point $N$. Prove that the length of segment $O N$ is a constant, and find this constant.
(1) $\frac{3}{2 a^{2}}+\frac{1}{4 b^{2}}=1 \Rightarrow \frac{3 b^{2}}{2 a^{2}}=b^{2}-\frac{1}{4} \Rightarrow \frac{3}{2}\left(1-e^{2}\right)=\frac{3}{4}=b^{2}-\frac{1}{4} \Rightarrow b^{2}=1, a^{2}=2$,

So the standard equation of the ellipse is $\frac{x^{2}}{2}+y^{2}=1$.
(2) Let the equation of the circle be $x^{2}+y^{2}-2 x-t y=0$, the center of the circle is
$\left(1, \frac{t}{2}\right)$, and the radius $r=\sqrt{1+\frac{t^{2}}{4}}$, then the distance from the center of the circle to the line is
$d=\frac{|3-2 t-5|}{5}=\frac{t}{2} \Rightarrow t=4$.
So the equation of the circle is $x^{2}+y^{2}-2 x-4 y=0$.
(3) The equation of the line perpendicular to point $F$ is $y=-\frac{2}{t}(x-1)$,

Combining $\left\{\begin{array}{l}x=-\frac{t}{2} y+1, \\ x^{2}+y^{2}-2 x-t y=0\end{array} \Rightarrow\left(\frac{t^{2}}{4}+1\right) y^{2}-t y-1=0\right.$.
Then $|O N|^{2}=x_{N}^{2}+y_{N}^{2}=\left(-\frac{t}{2} y_{N}+1\right)^{2}+y_{N}^{2}=\left(\frac{t^{2}}{4}+1\right) y_{N}^{2}-t y_{N}+1=2$, so the length of segment $O N$ is a constant $\sqrt{2}$.