If $\left(0, \frac{3}{4}\right)$ is the radical centre of the circles $S \equiv x^2+y^2+\alpha x+6 y=0, S \equiv x^2+y^2+2 \alpha x+\alpha y+6=0$ and $S^{\prime \prime} \equiv x^2+y^2+6 \alpha x-\alpha y+3=0$, then the distance between the radical centre and the centre of the circle $S^{\prime}=0$ is

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 $\frac{1}{2}\left(a^2 \beta^2+b^2 \alpha^2\right)=$