$A$ circle has the same center as an ellipse and passes through the foci $F_1$ and $F_2$ of the ellipse,such that the two curves intersect in $4$ points. Let $P$ be any one of their points of intersection. If the major axis of the ellipse is $17$ and the area of the triangle $PF_1F_2$ is $30$,then the distance between the foci is:

$A$ quadratic polynomial $y = f(x)$ with absolute term $3$ neither touches nor intersects the abscissa axis and is symmetric about the line $x = 1$. The coefficient of the leading term of the polynomial is unity. $A$ point $A(x_1, y_1)$ with abscissa $x_1 = 1$ and a point $B(x_2, y_2)$ with ordinate $y_2 = 11$ are given in a Cartesian rectangular system of coordinates $OXY$ in the first quadrant on the curve $y = f(x)$,where $O$ is the origin. The scalar product of the vectors $\vec{OA}$ and $\vec{OB}$ is:

If a tangent to the hyperbola $xy = -1$ is also a tangent to the parabola $y^2 = 8x$,then the equation of that tangent is

For $0 < \theta < \pi / 2$,if the eccentricity of the hyperbola $x^2 - y^2 \operatorname{cosec}^2 \theta = 5$ is $\sqrt{7}$ times the eccentricity of the ellipse $x^2 \operatorname{cosec}^2 \theta + y^2 = 5$,then the value of $\theta$ is:

An ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$ and the parabola $x^2 = 4(y + b)$ are such that the two foci of the ellipse and the end points of the latus rectum of the parabola are the vertices of a square. The eccentricity of the ellipse is

$PQ$ is a normal chord of the parabola $y^2 = 4ax$ at $P$,$A$ being the vertex of the parabola. Through $P$,a line is drawn parallel to $AQ$ meeting the $x$-axis in $R$. Then the length of $AR$ is: