Let the origin be the centre,$(\pm 3, 0)$ be the foci,and $\frac{3}{2}$ be the eccentricity of a hyperbola. Then the line $2x - y - 1 = 0$

The length of the conjugate axis of a hyperbola is greater than the length of the transverse axis. Then, the eccentricity $e$ is

Let $P(a \sec \theta, b \tan \theta)$ and $Q(a \sec \phi, b \tan \phi)$ where $\theta + \phi = \frac{\pi}{2}$ be two points on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. If $(h, k)$ is the point of intersection of the normals drawn at $P$ and $Q$,then $k=$

Let $A(2 \sec \theta, 3 \tan \theta)$ and $B(2 \sec \phi, 3 \tan \phi)$ where $\theta+\phi=\frac{\pi}{2}$ be two points on the hyperbola $\frac{x^2}{4}-\frac{y^2}{9}=1$. If $(\alpha, \beta)$ is the point of intersection of normals to the hyperbola at $A$ and $B$, then $\beta$ is equal to

If the area of the region bounded by $16x^2 - 9y^2 = 144$ and $8x - 3y = 24$ is $A$, then $3(A + 6 \ln(3))$ is equal to . . . . . . .

Find the equation of the hyperbola satisfying the given conditions: Vertices $(\pm 7, 0)$,$e = \frac{4}{3}$.