The slope of a common tangent to the ellipse $\frac{x^2}{49}+\frac{y^2}{4}=1$ and the circle $x^2+y^2=16$ is

Find the equation of the ellipse whose focus is $(-1, 1)$,eccentricity is $1/2$,and directrix is $x - y + 3 = 0$.

$A$ line perpendicular to the $X$-axis cuts the circle $x^2+y^2=9$ at $A$ and the ellipse $4x^2+9y^2=36$ at $B$ such that $A$ and $B$ lie in the same quadrant. If $\theta$ is the greatest acute angle between the tangents drawn to the curves at $A$ and $B$,then $\tan \theta=$

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:

$A$ 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:

Find the common tangent to the curves $x^2 + y^2 = 4$ and $2x^2 + y^2 = 2$.