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

The eccentricity of an ellipse $E$ with centre at the origin $O$ is $\frac{\sqrt{3}}{2}$ and its directrices are $x = \pm \frac{4\sqrt{6}}{3}$. Let $H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be a hyperbola whose eccentricity is equal to the length of semi-major axis of $E$, and whose length of latus rectum is equal to the length of minor axis of $E$. Then the distance between the foci of $H$ is :

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

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

The ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(b>a)$ and the parabola $y^2=4ax$ intersect at right angles. If $e$ is the eccentricity of the ellipse,then $2e^2=$

Let the circle $C$ touch the line $x - y + 1 = 0$,have the centre on the positive $x$-axis,and cut off a chord of length $\frac{4}{\sqrt{13}}$ along the line $-3x + 2y = 1$. Let $H$ be the hyperbola $\frac{x^2}{\alpha^2} - \frac{y^2}{\beta^2} = 1$,whose one of the foci is the centre of $C$ and the length of the transverse axis is the diameter of $C$. Then $2\alpha^2 + 3\beta^2$ is equal to . . . . . .