Find the coordinates of the foci and the vertices, the eccentricity,the length of the latus rectum of the hyperbolas : $y^{2}-16 x^{2}=16$
Dividing the equation by $16$ on both sides, we have $\frac{y^{2}}{16}-\frac{x^{2}}{1}=1$
Comparing the equation with the standard equation $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1,$ we find that $a=4, b=1$ and $c=\sqrt{a^{2}+b^{2}}=\sqrt{16+1}=\sqrt{17}$
Therefore, the coordinates of the foci are $(0, \,\pm \sqrt{17})$ and that of the vertices are $(0,\,\pm 4) .$ Also,
The eccentricity $e=\frac{c}{a}=\frac{\sqrt{17}}{4} .$
The latus rectum $=\frac{2 b^{2}}{a}=\frac{1}{2}$
Length of latus rectum of hyperbola $\frac{{{x^2}}}{{{{\cos }^2}\alpha }} - \frac{{{y^2}}}{{{{\sin }^2}\alpha }} = 4\,is\,\left( {\alpha \ne \frac{{n\pi }}{2},n \in I} \right)$
Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $16 x^{2}-9 y^{2}=576$
A hyperbola whose transverse axis is along the major axis of then conic, $\frac{{{x^2}}}{3} + \frac{{{y^2}}}{4} = 4$ and has vertices at the foci of this conic . If the eccentricity of the hyperbola is $\frac{3}{2}$ , then which of the following points does $NOT$ lie on it ?
Let $m_1$ and $m_2$ be the slopes of the tangents drawn from the point $P (4,1)$ to the hyperbola $H: \frac{y^2}{25}-\frac{x^2}{16}=1$. If $Q$ is the point from which the tangents drawn to $H$ have slopes $\left| m _1\right|$ and $\left| m _2\right|$ and they make positive intercepts $\alpha$ and $\beta$ on the $x$ axis, then $\frac{(P Q)^2}{\alpha \beta}$ is equal to $............$.
The distance between the directrices of a rectangular hyperbola is $10$ units, then distance between its foci is