Question

The parabola $y^2=4x$ and $x^2=4y$ divide the square region bounded by the lines $x=4,y=4$ and the coordinate axes. If $S_1,S_2,S_3$ are the areas of these parts numbered from top to bottom respectively, then

• A. $S_1:S_2\equiv 1:1$
• B. $S_2:S_3\equiv 1:2$
• C. $S_1:S_3\equiv 1:1$
• D. $S_1:(S_1+S_2)\equiv 1:2$

Answer 1

Answer: (A), (C), (D)

$S_1:S_2\equiv 1:1$
$S_1:S_3\equiv 1:1$
$S_1:(S_1+S_2)\equiv 1:2$

Area of region bounded by $y^2=4ax$ and $y^2=4by$ is $\frac{16ab}{3}$ sq. units.

Now $S_1=S_3={\int }_0^4\frac{x^2}{4}dx=\frac{1}{4}{\left[\frac{x^3}{3}\right]}_0^4=\frac{1}{12}\times 64=\frac{16}{3}$ sq. unit

$\therefore S_2+S_3={\int }_0^4\sqrt{4x}dx=2{\left[\frac{x^{3/2}}{3/2}\right]}_0^4=\frac{4}{3}\times 8=\frac{32}{3}$ sq. unit

$\Rightarrow S_2=\frac{16}{3}$ sq. unit

$\therefore S_1:S_2:S_3=\frac{16}{3}:\frac{16}{3}:\frac{16}{3}=1:1:1$