GUJCET 2010 Mathematics Question Paper with Answer and Solution

(A) The equation of the parabola is $y^2 = 8x$. Comparing this with $y^2 = 4ax$,we get $4a = 8$,which implies $a = 2$.
The focus of the parabola is $(a, 0) = (2, 0)$.
The equation of the latus rectum is $x = a = 2$.
The parabola is symmetric about the $x$-axis.
The area of the region bounded by the parabola and its latus rectum is given by $A = 2 \int_{0}^{2} y \, dx$.
Since $y^2 = 8x$,we have $y = \sqrt{8x} = 2\sqrt{2} \sqrt{x}$.
Substituting this into the integral,we get $A = 2 \int_{0}^{2} 2\sqrt{2} \sqrt{x} \, dx = 4\sqrt{2} \int_{0}^{2} x^{1/2} \, dx$.
Evaluating the integral: $A = 4\sqrt{2} \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{2} = 4\sqrt{2} \times \frac{2}{3} \times (2)^{3/2} = \frac{8\sqrt{2}}{3} \times 2\sqrt{2} = \frac{16 \times 2}{3} = \frac{32}{3}$ sq. units.