If $\frac{x^2}{\alpha+3}+\frac{y^2}{2-\alpha}=1$ represents a hyperbola,then $\alpha$ lies in

Statement $I$: The eccentricity of the hyperbola $9x^2-16y^2-72x+96y-144=0$ is $5/4$.
Statement $II$: The eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $\sqrt{1+\frac{b^2}{a^2}}$.

If the line $y = mx + 7\sqrt{3}$ is normal to the hyperbola $\frac{x^2}{24} - \frac{y^2}{18} = 1$,then a value of $m$ is

Let the $X$-axis be the transverse axis and the $Y$-axis be the conjugate axis of a hyperbola $H$. Let $x^2+y^2=16$ be the director circle of $H$. If the perpendicular distance from the centre of $H$ to its latus rectum is $\sqrt{34}$,then $a+b=$

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:
$(A) \left(\frac{9}{2\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
$(B) \left(-\frac{9}{2\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$
$(C) (3\sqrt{3}, -2\sqrt{2})$
$(D) (-3\sqrt{3}, 2\sqrt{2})$

The graph of the conic $x^2 - (y - 1)^2 = 1$ has one tangent line with positive slope that passes through the origin. The point of tangency is $(a, b)$. Then the eccentricity of the conic is: