\[ (4, 4\sqrt{3}) \text{ lies on } y^2 = 4ax \] \[ \Rightarrow 48 = 4a \cdot 4 \] \[ 4a = 12 \] \[ \Rightarrow y^2 = 12x \text{ is the equation of the parabola.} \] Now, the parameter for point \( P \) is \( t_1 = \frac{2}{\sqrt{3}} \). Therefore, the parameters for point \( Q \) are \( t_2 = -\frac{\sqrt{3}}{2} \), which gives the coordinates of \( Q \) as \( \left( \frac{9}{4}, -3\sqrt{3} \right) \). \textbf{Area of trapezium PQNM:} \[ \text{Area} = \frac{1}{2} \times MN \times (PM + QN) \] \[ = \frac{1}{2} \times MN \times (PS + QS) \] \[ = \frac{1}{2} \times MN \times PQ \] \[ = \frac{1}{2} \times 7\sqrt{3} \times \frac{49}{4} \] \[ = \frac{343\sqrt{3}}{8}\]
If the shortest distance of the parabola \(y^{2}=4x\) from the centre of the circle \(x² + y² - 4x - 16y + 64 = 0\) is d, then d2 is equal to:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)