The asymptotes of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$,with any tangent to the hyperbola form a triangle whose area is $a^2 \tan (\alpha)$. Then its eccentricity equals

At which point does the line $2x + \sqrt{6}y = 2$ touch the hyperbola $x^2 - 2y^2 = 4$?

If the product of the slopes of the tangents drawn from an external point $P(h, k)$ to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is a constant $k^2$,then the locus of $P$ is

If the straight line $(a-1)x - by + 4 = 0$ is normal to the hyperbola $xy = 1$, then which of the following does not hold?

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

Tangents are drawn to the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$,parallel to the straight line $2x-y=1$. The points of contact of the tangents on the hyperbola are:
$(A) \left(\frac{9}{2\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
$(B) \left(-\frac{9}{2\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$
$(C) (3\sqrt{3}, -2\sqrt{2})$
$(D) (-3\sqrt{3}, 2\sqrt{2})$