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?

Consider the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ having one of its foci at $P(-3,0)$. If the latus rectum through its other focus subtends a right angle at $P$ and $a^2b^2 = \alpha\sqrt{2} - \beta$,where $\alpha, \beta \in N$,then find the value of $\alpha + \beta$.

The equation of the normal at the point $(6, 4)$ on the hyperbola $\frac{x^2}{9} - \frac{y^2}{16} = 3$ is:

Let $S = \{(x,y) \in \mathbb{R}^2 : \frac{y^2}{1+r} - \frac{x^2}{1-r} = 1\}$,where $r \neq \pm 1$. Then $S$ represents

If ${m_1}$ and ${m_2}$ are the slopes of the tangents to the hyperbola $\frac{x^2}{25} - \frac{y^2}{16} = 1$ which pass through the point $(6, 2)$,then:

The equation of the pair of asymptotes of the hyperbola $4x^2 - 9y^2 - 24x - 36y - 36 = 0$ is