If the normal drawn to the hyperbola $xy=16$ at $(8,2)$ meets the hyperbola again at a point $(\alpha, \beta)$,then $|\beta|+\frac{1}{|\alpha|}=$

Let the points $P_1\left(\frac{\pi}{4}\right), P_2\left(\frac{3 \pi}{4}\right), P_3\left(\frac{5 \pi}{4}\right)$ and $P_4\left(\frac{7 \pi}{4}\right)$ given in parametric form,lie on the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=1$. Then these four points in that order form

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 $\frac{(\alpha_1+\alpha_2+\alpha_3+\alpha_4)}{(\beta_1+\beta_2+\beta_3+\beta_4)}(\alpha_1 \alpha_2 \alpha_3 \alpha_4) =$

The vertices of the hyperbola $7x^2 - 49y^2 = 343$ are

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

The product of the perpendiculars drawn from any point on a hyperbola to its asymptotes is