The equation of the hyperbola in the standard form (with transverse axis along the $x$ - axis) having the length of the latus rectum = $9$ units and eccentricity = $5/4$ is
$\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{{18}} = 1$
$\frac{{{x^2}}}{{36}} - \frac{{{y^2}}}{{27}} = 1$
$\frac{{{x^2}}}{{64}} - \frac{{{y^2}}}{{36}} = 1$
$\frac{{{x^2}}}{{36}} - \frac{{{y^2}}}{{64}} = 1$
The sound of a cannon firing is heard one second later at a position $B$ that at position $A$. If the speed of sound is uniform, then
Consider a branch of the hyperbola $x^2-2 y^2-2 \sqrt{2} x-4 \sqrt{2} y-6=0$ with vertex at the point $A$. Let $B$ be one of the end points of its latus rectum. If $\mathrm{C}$ is the focus of the hyperbola nearest to the point $\mathrm{A}$, then the area of the triangle $\mathrm{ABC}$ is
Tangents are drawn to the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$, parallel to the straight line $2 x-y=1$. The points of contacts of the tangents on the hyperbola are
$(A)$ $\left(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ $(B)$ $\left(-\frac{9}{2 \sqrt{2}},-\frac{1}{\sqrt{2}}\right)$
$(C)$ $(3 \sqrt{3},-2 \sqrt{2})$ $(D)$ $(-3 \sqrt{3}, 2 \sqrt{2})$
If the length of the transverse and conjugate axes of a hyperbola be $8$ and $6$ respectively, then the difference focal distances of any point of the hyperbola will be
The point $\mathrm{P}(-2 \sqrt{6}, \sqrt{3})$ lies on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ having eccentricity $\frac{\sqrt{5}}{2} .$ If the tangent and normal at $\mathrm{P}$ to the hyperbola intersect its conjugate axis at the point $\mathrm{Q}$ and $\mathrm{R}$ respectively, then $QR$ is equal to :