Consider a curve $$y=y(x)$$ in the first quadrant as shown in the figure. Let the area $$\mathrm{A}_{1}$$ is twice the area $$\mathrm{A}_{2}$$. Then the normal to the curve perpendicular to the line $$2 x-12 y=15$$ does NOT pass through the point.

The area of the smaller region enclosed by the curves $$y^{2}=8 x+4$$ and $$x^{2}+y^{2}+4 \sqrt{3} x-4=0$$ is equal to

The area bounded by the curves $$y=\left|x^{2}-1\right|$$ and $$y=1$$ is

The odd natural number a, such that the area of the region bounded by y = 1, y = 3, x = 0, x = y^a is $${{364} \over 3}$$, is equal to :