Find the locus of a point which moves such that its distance from the point $(0, 0)$ is twice its distance from the $y$-axis.

Through the vertex $O$ of the parabola $y^2 = 4ax$,two chords $OP$ and $OQ$ are drawn,and the circles on $OP$ and $OQ$ as diameters intersect in $R$. If $\theta_1, \theta_2$,and $\phi$ are the angles made with the axis by the tangents at $P$ and $Q$ on the parabola and by $OR$ respectively,then the value of $\cot \theta_1 + \cot \theta_2$ is:

The equation of the common tangent to the curves $y^2 = 8x$ and $xy = -1$ is:

Let $E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$ and $H : \frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$. Let the distance between the foci of $E$ and the foci of $H$ be $2\sqrt{3}$. If $a - A = 2$,and the ratio of the eccentricities of $E$ and $H$ is $\frac{1}{3}$,then the sum of the lengths of their latus rectums is equal to :

$A$ quadratic polynomial $y = f(x)$ with constant term $3$ neither touches nor intersects the $x$-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 vertex of the quadratic polynomial is:

The locus of the midpoints of the chords of the hyperbola $x^2 - y^2 = a^2$ which touch the parabola $y^2 = 4ax$ is: