The vertex and the focus of the parabola $2 x^2+5 y-6 x+1=0$ respectively, are

The axis of a parabola is along the line $y=x$ and the distance of its vertex $A$ from $(0,0)$ is $\sqrt{2}$ and that of its focus $S$ from $(0,0)$ is $2 \sqrt{2}$. If $A$ and $S$ lie in first quadrant, then the equation of the parabola in parametric form is

Let $S \equiv \frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0, S \equiv \frac{x^2}{\alpha^2}+\frac{y^2}{\beta^2}-1=0$ be two intersecting ellipses. If $P(a \cos \theta, b \sin \theta)$ and $Q\left(a \cos \left(\frac{\pi}{2}+\theta\right), b \sin \left(\frac{\pi}{2}+\theta\right)\right)$ are their points of intersection then $\frac{1}{2}\left(a^2 \beta^2+b^2 \alpha^2\right)=$

$P\left(\theta_1\right)$ and $Q\left(\theta_2\right)$ are two points on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with eccentricity $e$. If $P S Q$ is a focal chord and $\tan \left(\frac{\theta_1}{2}\right) \tan \left(\frac{\theta_2}{2}\right)=-(2 \sqrt{2}+3)$, then $e$ and $S$ are