If the eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ passing through the point $(4,6)$ is $2$,then the equation of the tangent to this hyperbola at $(4,6)$ is

Statement $ (A) $: The point $(5, -4)$ lies inside the hyperbola $y^2 - 9x^2 + 1 = 0$.
Reason $(R)$: $A$ point $(x_1, y_1)$ lies inside the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ if $\frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} - 1 < 0$.

$A$ 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

Let $a$ and $b$ be positive real numbers such that $a > 1$ and $b < a$. Let $P$ be a point in the first quadrant that lies on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Suppose the tangent to the hyperbola at $P$ passes through the point $(1, 0)$,and suppose the normal to the hyperbola at $P$ cuts off equal intercepts on the coordinate axes. Let $\Delta$ denote the area of the triangle formed by the tangent at $P$,the normal at $P$,and the $x$-axis. If $e$ denotes the eccentricity of the hyperbola,then which of the following statements is/are $TRUE$?
$(A)$ $1 < e < \sqrt{2}$
$(B)$ $\sqrt{2} < e < 2$
$(C)$ $\Delta = a^4$
$(D)$ $\Delta = b^4$

Find the coordinates of the foci and the vertices,the eccentricity,and the length of the latus rectum of the hyperbola: $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$.

The equation of the normal to the hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$ at the point $(8, 3\sqrt{3})$ is