Let the foci of a hyperbola $$H$$ 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 $$H$$ be the reciprocal of the eccentricity of the ellipse $$E$$. If the length of the transverse axis of $$H$$ is $$\alpha$$ and the length of its conjugate axis is $$\beta$$, then $$3 \alpha^2+2 \beta^2$$ is equal to
Let $$H: \frac{-x^2}{a^2}+\frac{y^2}{b^2}=1$$ be the hyperbola, whose eccentricity is $$\sqrt{3}$$ and the length of the latus rectum is $$4 \sqrt{3}$$. Suppose the point $$(\alpha, 6), \alpha>0$$ lies on $$H$$. If $$\beta$$ is the product of the focal distances of the point $$(\alpha, 6)$$, then $$\alpha^2+\beta$$ is equal to
Consider a hyperbola $$\mathrm{H}$$ having centre at the origin and foci on the $$\mathrm{x}$$-axis. Let $$\mathrm{C}_1$$ be the circle touching the hyperbola $$\mathrm{H}$$ and having the centre at the origin. Let $$\mathrm{C}_2$$ be the circle touching the hyperbola $$\mathrm{H}$$ at its vertex and having the centre at one of its foci. If areas (in sq units) of $$C_1$$ and $$C_2$$ are $$36 \pi$$ and $$4 \pi$$, respectively, then the length (in units) of latus rectum of $$\mathrm{H}$$ is
$x^2-y^2 \operatorname{cosec}^2 \theta=5$ is $\sqrt{7}$ times eccentricity of the
ellipse $x^2 \operatorname{cosec}^2 \theta+y^2=5$, then the value of $\theta$ is :