The equation of the director circle of the hyperbola $\frac{x^2}{16} - \frac{y^2}{4} = 1$ is given by

$P$ is a point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. $N$ is the foot of the perpendicular from $P$ on the transverse axis. The tangent to the hyperbola at $P$ meets the transverse axis at $T$. If $O$ is the centre of the hyperbola,then $OT \cdot ON$ is equal to:

Find the eccentricity of the hyperbola $16x^{2} - 32x - 3y^{2} + 12y = 44$.

Let $x$ be the eccentricity of a hyperbola whose transverse axis is twice its conjugate axis. Let $y$ be the eccentricity of another hyperbola for which the distance between the foci is $3$ times the distance between its directrices. Then $y^2-x^2=$

The foci of the hyperbola $\frac{x^2}{16} - \frac{(y - 2)^2}{9} = 1$ are:

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: