Eccentricity of conjugate hyperbola of $16x^2 – 9y^2 – 32x – 36y – 164 = 0$ will be-

If the eccentricity of the hyperbola $x^2 – y^2 \sec^2 \alpha = 5$  is $\sqrt 3 $  times the eccentricity of the ellipse $x^2 \sec^2 \alpha + y^2 = 25, $ then a value of $\alpha$  is :

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 :

If a directrix of a hyperbola centered at the origin and passing through the point $(4, -2\sqrt 3)$ is $5x = 4\sqrt 5$ and its eccentricity is $e$, then

The eccentricity of the conjugate hyperbola of the hyperbola ${x^2} – 3{y^2} = 1$, is