If the equation of the hyperbola having $(8,3)$ and $(0,3)$ as foci and $\frac{4}{3}$ as eccentricity is $\frac{(x-\alpha)^2}{p}-\frac{(y-\beta)^2}{q}=1$,then $p+q=$

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 $xy = c^2$,the coordinates of the orthocentre of the triangle $PQR$ are:

If the normals drawn to the hyperbola $xy=4$ at $(\alpha_i, \beta_i)$ for $i=1, 2, 3, 4$ are concurrent at the point $(a, b)$,then $\frac{(\alpha_1+\alpha_2+\alpha_3+\alpha_4)}{(\beta_1+\beta_2+\beta_3+\beta_4)}(\alpha_1 \alpha_2 \alpha_3 \alpha_4) =$

$P$ 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:

If the product of the perpendicular distances from any point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ of eccentricity $e = \sqrt{3}$ from its asymptotes is equal to $6$,then the length of the transverse axis of the hyperbola is

$A$ hyperbola passes through a focus of the ellipse $\frac{x^2}{169}+\frac{y^2}{25}=1$. Its transverse and conjugate axes coincide respectively with the major and minor axes of the ellipse. The product of their eccentricities is $1$. Then,the equation of the hyperbola is