The equation of the hyperbola whose directrix is $x + 2y = 1$, focus $(2, 1)$ and eccentricity $2$ will be

What will be equation of that chord of hyperbola $25{x^2} - 16{y^2} = 400$, whose mid point is $(5, 3)$

Let the foci of a hyperbola $\mathrm{H}$ coincide with the foci of the ellipse $E: \frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1$ and the eccentricity of the hyperbola $\mathrm{H}$ be the reciprocal of the eccentricity of the ellipse $E$. If the length of the transverse axis of $\mathrm{H}$ is $\alpha$ and the length of its conjugate axis is $\beta$, then $3 \alpha^2+2 \beta^2$ is equal to :

The distance between the directrices of a rectangular hyperbola is $10$ units, then distance between its foci is

If $ PN$  is the perpendicular from a point on a rectangular hyperbola $x^2 - y^2 = a^2 $ on any of its asymptotes, then the locus of the mid point of $PN$  is :

Let $e_1$ be the eccentricity of the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ and $e_2$ be the eccentricity of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$, which passes through the foci of the hyperbola. If $e_1 e_2=1$, then the length of the chord of the ellipse parallel to the $\mathrm{x}$-axis and passing through $(0,2)$ is :