Well I already got the answer to that and it is $e =\sqrt{ \frac{2\sqrt{4h^2 +(a-b)^2}}{a+b + \sqrt{4h^2 +(a-b)^2}}}$
But I proved it for an ellipse, surprisingly it works for all other conics too!
I have posted my solution (https://math.stackexchange.com/a/4215136/955104) and I would like to know why it works for all the other conics.
Given: we need to find the eccentricity of ellipse $ \ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 $
[Because it is ellipse, $\ h^2 \le ab $, so consider $\ a,b \gt 0$. ]
Now we know that centre of ellipse is $ \left( \frac{hf-bg}{ab-h^2},\frac{hg-af}{ab-h^2} \right)=\left(\alpha, \beta\right)$ (Let's say).
Note that $(\alpha,\beta)$ is solution of the lines $ax+hy+g=0 ,\ hx+by+f=0$, which implies $\boxed { a\alpha+h\beta+g=0 , h\alpha+b\beta+f=0 }\tag{1}$
Now let P be a point on ellipse at distance $r$ from centre then P is in the form$\left( \alpha +r\cos \theta , \beta +r\sin \theta\right)$
Before substituting P in ellipse equation let's find the value of
\begin{align} E &=a(x_1+x_2)^2 + 2h(x_1+x_2)(y_1+y_2) + b(y_1+y_2)^2 \\ &\quad +2g(x_1+x_2) + 2f(y_1+y_2) + c \\ \implies E&= (ax_1^2 + 2hx_1y_1 + by_1^2 + 2gx_1 + 2fy_1 +c) \\&\quad+ (ax_2^2 + 2hx_2y_2 + by_2^2 + 2gx_2 + 2fy_2) + 2ax_1x_2 \\&\quad +2h(x_1y_2+x_2y_1)+2by_1y_2 \tag{2} \end{align} Now using (2) let's put P in ellipse equation. We get
$$ (a\alpha^2 + 2h\alpha \beta + b\beta^2 + 2g\alpha + 2f\beta + c) + r^2(a\cos^2\theta + 2h\cos\theta \sin \theta + b\sin^2 \theta) + 2gr\cos\theta + 2fr\sin \theta +2a \alpha r \cos\theta + 2hr(\alpha\sin\theta + \beta \cos \theta ) + 2b\beta r \sin\theta = 0 $$ Let $-\Delta = (a\alpha^2 + 2h\alpha \beta + b\beta^2 + 2g\alpha + 2f\beta + c)$. Then $$ -\Delta + r^2(a\cos^2\theta + 2h\cos\theta \sin \theta + b\sin^2 \theta) + 2r\cos\theta( a\alpha+h\beta+g)+2r\sin\theta(h\alpha+b\beta+f) = 0 $$
From (1) we get
\begin{align} &-\Delta + r^2(a\cos^2\theta + 2h\cos\theta \sin \theta + b\sin^2 \theta) = 0 \\ \implies &r^2 = \frac{\Delta}{a\cos^2\theta + 2h\cos\theta \sin \theta + b\sin^2 \theta} \\ \implies &r^2 = \frac{2\Delta}{a(2\cos^2\theta) + 2h(2\cos\theta \sin \theta) + b(2\sin^2 \theta)} \\ &\quad [\because 2\sin\theta\cos\theta=\sin2\theta , 2\cos^2\theta= 1+\cos2\theta, 2\sin^2\theta = 1 -cos2\theta ] \\ \implies &r^2 = \frac{2\Delta}{a(1+cos2\theta) + 2h(\sin 2\theta) + b(1-\cos2 \theta)} \\ \implies &r^2 = \frac{2\Delta}{(a+b)+2h\sin2\theta+(a-b)\cos2\theta} \end{align}
Now
\begin{align} &2h\sin2\theta+(a-b)\cos2\theta \in[-\sqrt{4h^2+(a-b)^2},+\sqrt{4h^2+(a-b)^2}] \\ &\implies a+b+2h\sin2\theta+(a-b)\cos2\theta \\&\qquad \qquad \in\left[a+b-\sqrt{4h^2+(a-b)^2},a+b+\sqrt{4h^2+(a-b)^2}\right] \end{align} Note that since equation is ellipse, $$\boxed { \ h^2 \lt ab \\ \implies 4h^2 +(a-b)^2 \lt (a-b)^2 +4ab = (a+b)^2 \\ \implies 4h^2 +(a-b)^2 \lt (a+b)^2 \\ \implies \sqrt{4h^2 +(a-b)^2} \lt a+b \ [\because a,b \gt 0] \\ \implies a+b - \sqrt{4h^2 +(a-b)^2} \gt 0 } \\ \implies \frac{2\Delta}{r^2} \in \left[a+b-\sqrt{4h^2+(a-b)^2},a+b+\sqrt{4h^2+(a-b)^2}\right] \\ \implies r^2 \in \left[\frac{2\Delta}{a+b+\sqrt{4h^2+(a-b)^2}},\frac{2\Delta}{a+b-\sqrt{4h^2-(a-b)^2}}\right] $$
Now
$$ r_{max}^2 = (\text{semi major axis})^2 = M^2 \ [\text{Let's say}] \\ r_{min}^2 = (\text{semi minor axis})^2 = m^2 \ [\text{Let's say}] $$
Now
\begin{align} e^2 &= 1-\frac{m^2}{M^2} = 1 - \frac{a+b - \sqrt{4h^2 +(a-b)^2}}{a+b + \sqrt{4h^2 +(a-b)^2}} \\ \implies e^2 &= \frac{2\sqrt{4h^2 +(a-b)^2}}{a+b + \sqrt{4h^2 +(a-b)^2}} \\ \implies e &=\sqrt{ \frac{2\sqrt{4h^2 +(a-b)^2}}{a+b + \sqrt{4h^2 +(a-b)^2}}} \end{align} $$\therefore \boxed{e =\sqrt{ \frac{2\sqrt{4h^2 +(a-b)^2}}{a+b + \sqrt{4h^2 +(a-b)^2}}}} $$