Step 1: The equation of a parabola with focus \( (h, k) \) and directrix \( y = k - p \) is given by: \[ \frac{(x - h)^2}{4p} = y - k. \] For the first parabola with focus \( (4, 3) \) and directrix as the x-axis, we substitute the values \( h = 4 \), \( k = 3 \), and the directrix to get the equation of the first parabola.
Step 2: Similarly, for the second parabola with the same focus \( (4, 3) \) and the y-axis as the directrix, we can derive the equation of this parabola as well.
Step 3: To find the points of intersection, solve the two equations simultaneously. The distance between the intersection points A and B is \( AB \), and then we compute \( (AB)^2 \). After solving, the value of \( (AB)^2 \) is found to be 392. Thus, the correct answer is (1).
A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following: 
(i) Express \(y\) as a function of \(x\) from the given equation of ellipse.
(ii) Integrate the function obtained in (i) with respect to \(x\).
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.
OR
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\).
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.
Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________.
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]