$A$ point moves in such a way that the difference of its distances from two points $(8,0)$ and $(-8,0)$ always remains $4$. Then, the locus of the point is

$A$ normal to the hyperbola $4x^2 - 9y^2 = 36$ meets the coordinate axes $x$ and $y$ at $A$ and $B$,respectively. If the parallelogram $OABP$ ($O$ being the origin) is formed,then the locus of $P$ is

The normal to the rectangular hyperbola $xy = c^{2}$ at the point $t$ meets the curve again at a point $t'$,such that

The tangent to the hyperbola $xy = c^2$ at the point $P(ct, c/t)$ intersects the $x$-axis at $T$ and the $y$-axis at $T'$. The normal to the hyperbola at $P$ intersects the $x$-axis at $N$ and the $y$-axis at $N'$. If the areas of the triangles $PNT$ and $PN'T'$ are $\Delta$ and $\Delta'$ respectively,then $\frac{1}{\Delta} + \frac{1}{\Delta'}$ is:

If the length of the transverse and conjugate axes of a hyperbola are $8$ and $6$ respectively,then the difference of the focal distances of any point on the hyperbola will be

If the tangents drawn to the hyperbola $4y^2 = x^2 + 1$ intersect the coordinate axes at the distinct points $A$ and $B$,then the locus of the midpoint of $AB$ is