Let the line $y-x=1$ intersect the ellipse $\frac{x^2}{2}+\frac{y^2}{1}=1$ at the points A and B . Then the angle made by the line segment AB at the center of the ellipse is :

Let S and $\mathrm{S}^{\prime}$ be the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ and $\mathrm{P}(\alpha, \beta)$ be a point on the ellipse in the first quadrant. If $(\mathrm{SP})^2+\left(\mathrm{S}^{\prime} \mathrm{P}\right)^2-\mathrm{SP} \cdot \mathrm{S}^{\prime} \mathrm{P}=37$, then $\alpha^2+\beta^2$ is equal to :

If the line $\alpha x+4 y=\sqrt{7}$, where $\alpha \in \mathbf{R}$, touches the ellipse $3 x^2+4 y^2=1$ at the point P in the first quadrant, then one of the focal distances of $P$ is :

Let the ellipse $3x^2 + py^2 = 4$ pass through the centre $C$ of the circle $x^2 + y^2 - 2x - 4y - 11 = 0$ of radius $r$. Let $f_1, f_2$ be the focal distances of the point $C$ on the ellipse. Then $6f_1f_2 - r$ is equal to