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):\( (\sqrt{5},-2)\)
The heat generated in 1 minute between points A and B in the given circuit, when a battery of 9 V with internal resistance of 1 \(\Omega\) is connected across these points is ______ J. 
The given circuit works as: 
Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}
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.
