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

An ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ passes through the vertices of the hyperbola $H: \frac{x^{2}}{49}-\frac{y^{2}}{64}=-1$. The major and minor axes of the ellipse $E$ coincide with the transverse and conjugate axes of the hyperbola $H$. Let the product of the eccentricities of $E$ and $H$ be $\frac{1}{2}$. If $l$ is the length of the latus rectum of the ellipse $E$,then the value of $113l$ is equal to $....$

An ellipse has eccentricity $\frac{1}{2}$ and one focus at the point $P\left( \frac{1}{2}, 1 \right)$. Its one directrix is the common tangent nearer to the point $P$, to the circle $x^2 + y^2 = 1$ and the hyperbola $x^2 - y^2 = 1$. The equation of the ellipse in the standard form is:

Suppose that the foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1$ are $(f_1, 0)$ and $(f_2, 0)$ where $f_1 > 0$ and $f_2 < 0$. Let $P_1$ and $P_2$ be two parabolas with a common vertex at $(0,0)$ and with foci at $(f_1, 0)$ and $(2f_2, 0)$,respectively. Let $T_1$ be a tangent to $P_1$ which passes through $(2f_2, 0)$ and $T_2$ be a tangent to $P_2$ which passes through $(f_1, 0)$. If $m_1$ is the slope of $T_1$ and $m_2$ is the slope of $T_2$,then the value of $(\frac{1}{m_1^2} + m_2^2)$ is

The equations of the common tangents to the ellipse $x^2 + 4y^2 = 8$ and the parabola $y^2 = 4x$ are

The area of the quadrilateral formed by the foci of the hyperbolas $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $\frac{x^2}{a^2} - \frac{y^2}{b^2} = -1$ is