Interesting Parabola

$x^2=4y$ represents a parabola whose vertex is (0,0) and focus (0,0). This parabola is symmetric about co-ordinate axes. Area is bounded between lines $y=mx+c$ and parabola for $c>0$ . But for same value of $c$ , the area bounded by horizontal lines is minimum , because if there are lines $y=a$ and $y=mx+a$ , where m can be any integer , m≠0. then it can be assumed that the horizontal line is rotated by angle $tan^(-1)⁡m$ through (0,a) , either clockwise or anticlockwise, depending upon $m$ . hence on one side line moves upwards and on other side line moves downward . As the line moves upwards on one side , so its intersecting $x$ co-ordinate with parabola increases and on other side its it decreases. So area bounded between line and curve increases on one side and decreases on other side. But this increase and decrease doesnot nullify each other. Because its a parabola and not a
regular shape. So when its intersecting $x$ co-ordinate increases , so slope at each point also increases and when its intersecting $x$ co-ordinate decreases, slope at each point also decreases. Hence the increase in area is much more than decrease in area . So total area of parabola increases. This proves that area bounded by horizontal line for constant intercept on Y-axis is minimum. So at maximum value of , $'c'$ , intecept on Y-axis , the region bounded by $y=c$ and parabola is $''8''$ . So,

$\Large 2 (∫_0^cxdy)= 2(∫_0^c2√y)= 8,$

$\Large 8/3 c^\frac{3}{2} = 8$

$\Large c= 9 \frac{1}{3}$

Hence $\Large c ≤ 9 \frac{1}{3}$

$\Large∴ c ≤ 2$