If the line x – 1 = 0 is a directrix of the hyperbola kx2 – y2 = 6, then the hyperbola passes through the point
\((-2\sqrt5,6)\)
\((-\sqrt5,3)\)
\((\sqrt5,-2)\)
\((2\sqrt5,3\sqrt6)\)
Given hyperbola : \(\frac{x^2}{6/k}-\frac{y^2}{6} = 1\)
\(e = \sqrt{1+\frac{6}{6/k}}=\sqrt{1+k}\)
\(x = ± \frac{a}{e} ⇒ x = ± \frac{\sqrt6}{\sqrt{k}\sqrt{k+1}}\)
As given : \(\frac{\sqrt6}{\sqrt{k}\sqrt{k+1}}=1\)
\(⇒ k = 2\)
\(⇒ \frac{x^2}{3}-\frac{y^2}{6} = 1\)
Hence, the option that satisfies and is the correct option is (C):\( (\sqrt{5},-2)\)

Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).

Hyperbola is the locus of all the points in a plane such that the difference in their distances from two fixed points in the plane is constant.
Hyperbola is made up of two similar curves that resemble a parabola. Hyperbola has two fixed points which can be shown in the picture, are known as foci or focus. When we join the foci or focus using a line segment then its midpoint gives us centre. Hence, this line segment is known as the transverse axis.
