Let the eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be reciprocal to that of the ellips $x^2+4 y^2=4$. If the hyperbola passes through a focus of the ellipse, then

$(A)$ the equation of the hyperbola is $\frac{x^2}{3}-\frac{y^2}{2}=1$

$(B)$ a focus of the hyperbola is $(2,0)$

$(C)$ the eccentricity of the hyperbola is $\sqrt{\frac{5}{3}}$

$(D)$ the equation of the hyperbola is $x^2-3 y^2=3$

If $e$ and $e’$ are eccentricities of hyperbola and its conjugate respectively, then

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 locus of the point of intersection of the lines $(\sqrt{3}) kx + ky -4 \sqrt{3}=0$ and $\sqrt{3} x-y-4(\sqrt{3}) k=0$ is a conic, whose eccentricity is ………….

If $\mathrm{e}_{1}$ and $\mathrm{e}_{2}$ are the eccentricities of the ellipse, $\frac{\mathrm{x}^{2}}{18}+\frac{\mathrm{y}^{2}}{4}=1$ and the hyperbola, $\frac{\mathrm{x}^{2}}{9}-\frac{\mathrm{y}^{2}}{4}=1$ respectively and $\left(\mathrm{e}_{1}, \mathrm{e}_{2}\right)$ is a point on the ellipse, $15 \mathrm{x}^{2}+3 \mathrm{y}^{2}=\mathrm{k},$ then $\mathrm{k}$ is equal to