The length of the latus rectum of the hyperbola $25x^2 -16y^2 = 400$ is -
$\frac{25}{2}$
$\frac{16}{5}$
$\frac{15}{4}$
$\frac{4}{5}$
If ${m_1}$ and ${m_2}$are the slopes of the tangents to the hyperbola $\frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{{16}} = 1$ which pass through the point $(6, 2)$, then
The auxiliary equation of circle of hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$, is
The length of the latus rectum and directrices of a hyperbola with eccentricity e are 9 and $\mathrm{x}= \pm \frac{4}{\sqrt{3}}$, respectively. Let the line $y-\sqrt{3} \mathrm{x}+\sqrt{3}=0$ touch this hyperbola at $\left(\mathrm{x}_0, \mathrm{y}_0\right)$. If $\mathrm{m}$ is the product of the focal distances of the point $\left(\mathrm{x}_0, \mathrm{y}_0\right)$, then $4 \mathrm{e}^2+\mathrm{m}$ is equal to ...........
Centre of hyperbola $9{x^2} - 16{y^2} + 18x + 32y - 151 = 0$ is
If $ P(x_1, y_1), Q(x_2, y_2), R(x_3, y_3) $ and $ S(x_4, y_4) $ are $4 $ concyclic points on the rectangular hyperbola $x y = c^2$ , the co-ordinates of the orthocentre of the triangle $ PQR$ are :