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 $=\frac{2 b^{2}}{a}=\frac{1}{2}$
The distance between the directrices of the hyperbola $x = 8\sec \theta ,\;\;y = 8\tan \theta $ is
A line parallel to the straight line $2 x-y=0$ is tangent to the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{2}=1$ at the point $\left(x_{1}, y_{1}\right) .$ Then $x_{1}^{2}+5 y_{1}^{2}$ is equal to
The equation of the tangent to the conic ${x^2} - {y^2} - 8x + 2y + 11 = 0$ at $(2, 1)$ is
The line $y = mx + c$ touches the curve $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$, if
Let the eccentricity of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is reciprocal to that of the hyperbola $2 x^2-2 y^2=1$. If the ellipse intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is $................$.